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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1801. |
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply. The power consumed will beA. `750` wattB. `500` wattC. `250` wattD. `1000` watt |
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Answer» Correct Answer - C `R=(V^(2))/(P)=((220)^(2))/(1000)` Where `V` and `P` are denoting rated voltage and power respectively. `P_("consumed")=(V^(2))/R=(110xx110)/(220xx220)xx1000=250` watt |
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| 1802. |
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply. The power consumed will beA. 750 wattB. 500 wattC. 250 wattD. 1000 watt |
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Answer» Correct Answer - C (c) We know that R = `V_(rated)^(2)/P_(rated) = (220)^2/(1000)` When this bulb is connected to 110 volt mains supply we get `P = V^2/R = ((110)^2 xx 1000)/(220)^2 = 1000/4 = 250W` . |
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| 1803. |
Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell isA. 2.1 VB. 2.7 VC. 3 VD. 4.2 V |
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Answer» Correct Answer - A `r_(1)=0.2 Omega, r_(2)=0.4 Omega` V=lR `R=(r_(1)r_(2))/(r_(1)+r_(2)), l=l_(1)+l_(2) implies V=(r_(1)r_(2))/(r_(1)+r_(2))(l_(1)+l_(2))` `l_(1)=E_(1)//r_(1) " and " l_(2)=E_(2)//r_(2)` `V=((E_(1))/(r_(1))+(E_(2))/(r_(2)))(r_(1)r_(2))/(r_(1)+r_(2))=((E_(2)r_(1)+E_(1)r_(2)))/(r_(1)r_(2))((r_(1)r_(2))/(r_(1)+r_(2)))` ` V=(E_(2)r_(1)+E_(1)r_(2))/(r_(1)+r_(2)) implies 1.5=(E_(2)xx0.21.2xx0.4)/(0.6)` `0.9=0.2 E_(2)+0.48 implies (0.42)/(0.2)=E_(2) implies E_(2)=2.1 V` |
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| 1804. |
A battery of emf 2 V and internal resistance `0.5 Omega` is connected across a resistance of `9.5 Omega`. How many electrons pass through a cross-section of the resistance in 1 second ? |
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Answer» Correct Answer - `1.25xx10^(18)` |
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| 1805. |
If in the circuit shown below, the internal resistance of the battery is `1.5 Omega` and `V_(P)` and `V_(Q)` are the potential at `P` and `Q` respectively, what is the potential difference between the point `P` and `Q`? A. ZeroB. 4 volts `(V gt V)`C. 4 volts `(V gt V)`D. 2.5 volts `(V gt V)` |
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Answer» Correct Answer - D |
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| 1806. |
If in the circuit shown below, the internal resistance of the battery is `1.5 Omega` and `V_(P)` and `V_(Q)` are the potential at `P` and `Q` respectively, what is the potential difference between the point `P` and `Q`? A. ZeroB. `4 V(V_(P) gt V_(Q))`C. `4 V (V_(Q) gt V_(P))`D. `2.5 V V(V_(Q) gt V_(P))` |
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Answer» Correct Answer - D Effective resistance of circuit `=(5xx5)/(5+5)+15=4Omega` Total current in circuit, `I=(20)/(4)=5A` Current in arm APB or AQB`=(1)/(2)=2.5A` `V_(A)-V_(P)=3xx2.5=7.5V` `V_(A)-V_(Q)=2xx2.5=5V` `V_(P)-V_(Q)=(V_(A)-V_(Q))-(V_(A)-V_(P))` `=5.0-7.5=-2.5V` |
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| 1807. |
The current in conductor varies with time `t` as `I = 2 t + 3 t^(2)` where `I` is in ampere and `t` in seconds. Electric charge flowing through a section of the conductor during `t = 2 sec` to `t = 3 sec` isA. `10 C`B. `24 C`C. `33 C`D. `44 C` |
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Answer» Correct Answer - B (b) `dQ = Idt implies Q = int_(t = 2)^(t = 3) Idt = [ 2 int_(2)^(3) tdt + 3 int_(2)^(3) t^(2) dt]` `= [t^(2)]_(2)^(3) + [t^(3)]_(2)^(3) = (9 - 4) + (27 - 8) = 5 + 19 = 24 C`. |
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| 1808. |
A torch bulb rated `4.5 W, 1.5 V` is connected as shown in Fig. 7.35. The emf of the cell needed to make the bulb glow at full intensity is A. `4.5 V`B. `1.5 V`C. `2.67 V`D. `13.5 V` |
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Answer» Correct Answer - D (d) Current in the bulb `= (P).(V) = (4.5)/(1.5) = 3 A` Current in `1 Omega` resistance `= (1.5)/(1) = 1.5 A` Hence total current from the cell `i= 3 + 1.5 = 4.5 A` By using `E = V + ir implies E 1.5 + 4.5 xx (2.67) = 13.5 V)` |
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| 1809. |
If a potential difference of 1 volt applied across the conductor causes a current of 1 ampere to flow through, it then the resistance of a conductor is aA. ohmB. siemenC. faradD. henry |
| Answer» Correct Answer - A | |
| 1810. |
Potential differences across the terminals of a cell were measured (in volt) against different currents (in ampere) flowing through the cell. A graph was drawn which was a straight line ABC as shown in figure Determine from graph (i) emf of the cell (ii) maximum current obtained from the cell and (iii) internal resistance of the cell. |
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Answer» (i) EMF of the cell is equal to maximum potential difference across the two electrodes of cell corresponding to zero current. Thus emf of the cell, `epsilon = 1.4 V`. (ii) Max. current is drawn from the cell when the terminal pot. Diff. is zero. Therefore `I_(max) = 0.28 A` (iii) Internal resistance, `r=epsilon/I_(max) = (1.4 V)/(0.28A)= 5 Omega` |
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| 1811. |
A torch bulb rated `4.5 W, 1.5 V` is connected as shown in Fig. 7.35. The emf of the cell needed to make the bulb glow at full intensity is A. `4.5 V`B. `1.5 V`C. `2.67 V`D. `13.5V` |
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Answer» Correct Answer - D |
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| 1812. |
The e.m.f. of a cell is 6 volt. When 2 ampere current is drawn from it then th e potential difference across its terminal remains 3 volt. Its internal resistance. |
| Answer» Internal resistance `r=(E-V)/( I)=(6-3)/(2)=1.5Omega` | |
| 1813. |
The resistance of a wire is `10^(-6)Omega` per metre. It is bend in the form of a circle of diameter m 2 . A wire of the same material is connected across its diameter. The total resistance across its diameter AB will be A. `(4)/(3)pi xx 10^(-6)Omega`B. `(2)/(3) pi xx 10^(-6)Omega`C. `0.88 xx 10^(-6) Omega`D. `14 pi xx 10^(-6) Omega` |
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Answer» Correct Answer - C |
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| 1814. |
When the key `E` is pressed at time `t = 0`, which of the following statements about the current `I` in the resistor `AB` of the given circuit is true. A. `l = 2mA` at all tB. l oscillates between 1 mA and 2mAC. `l = 1 mA` at all tD. At `t = 0, 1 = 2 mA` and with time it goes to 1mA |
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Answer» Correct Answer - D |
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| 1815. |
When the key `E` is pressed at time `t = 0`, which of the following statements about the current `I` in the resistor `AB` of the given circuit is true. A. `I = 2 mA` at all `t`B. `I` oscillates between `1 mA` and `2 mA`C. `I = 1 mA` at all `t`D. At `t = 0, I = 2 mA` and with time it goes to `1 mA` |
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Answer» Correct Answer - D (d) At time `t = 0` i.e., when capacitor is charging current `i= (2)/(1000) = 2 mA` When capacitor is full charged, no current will pass through it, hence current through the circuit `i= (2)/(2000) = 1 mA` |
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| 1816. |
In the figure shown, the capacity of the consider `C` is `2 nu F`. The current in `2 Omega` resistor is A. 9AB. 0.9AC. `(1)/(9)A`D. `(1)/(0.9)A` |
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Answer» Correct Answer - B |
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| 1817. |
In the figure shown, the capacity of the consider `C` is `2 nu F`. The current in `2 Omega` resistor is A. `9 A`B. `0.9 A`C. `(1)/(9) A` and `9 V`D. `(1)/(0.9) A` |
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Answer» Correct Answer - B (b) No current flows through the capacitor branch in steady state. Total supplied by the battery. `i= (6)/(2.8 + 1.2) = (3)/(2)`, Current throught `2 Omega` resistor `= (3)/(2) xx (3)/(2) = 0.9A` |
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| 1818. |
In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of `2Omega`, the balancing length becomes 120 cm.The internal resistance of the cell isA. `4 Omega`B. `2 Omega`C. `1 Omega`D. `0.5 Omega` |
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Answer» Correct Answer - B The internal resistance of the cell. `r=((l_(1)-l_(2))/l_(2))R` `=(240-120)/120xx2=2 Omega` |
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| 1819. |
In the figure shown, the capacity of the consider `C` is `2 nu F`. The current in `2 Omega` resistor is A. `9 A`B. `0.9 A`C. `(1)/(9) A`D. `(1)/(0.9) A` |
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Answer» Correct Answer - B (b) No current flows through the capacitor barach on steady state. Total current supplied by the battery `i= (6)/(2.8 + 1.2) = (3)/(2)` Current through `2 Omega` resistor `= (3)/(2) xx (3)/(5) = 0.9 A` |
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| 1820. |
Two sources of current of equal emf are connected in series and having different internal resistance `r_1` and `r_2(r_2gtr_1)`. Find the external resistance `R` at which the potential difference across the terminals of one of the sources becomes equal to zero. |
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Answer» `V=E-ir` `E` and `i` for both the sources are equal. Therefore, potential difference (`V`) will be zero for a source having greater internal resistance i.e. `r_2` `:. 0=E-ir_2` or `E=ir_2=((2E)/(R+r_1_r_2)).r_2` `:. 2r_2=R+r_1+r_2` or `R=r_2-r_1` |
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| 1821. |
Two bulbs of (40W 200V) and (100W 200V). Then correct relation for their resistances:A. `R_(40)ltR_(100)`B. `R_(40)gtR_(100)`C. `R_(40)=R_(100)`D. no relation can be predicted. |
| Answer» `P=(V^(2))/(R),Pprop(1)/(R)` i.e. `R_(40)gtR_(100)` | |
| 1822. |
A combination of four resistances is shown in Fig. 4.59. Calculate the potential difference between the points P and Q, and the values of currents flowing in the different resistances. |
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Answer» Correct Answer - `14.4 V, 0.8 A, 1.6 A` |
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| 1823. |
Two resistors of resistances `2Omega` and `6Omega` are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance `0.5 Omega`. What is the current flowing thrpough the battery ?A. `4 A`B. `(4)/(3)A`C. `(4)/(17)A`D. `1 A` |
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Answer» Correct Answer - D |
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| 1824. |
In the series combination of two or more than two resistancesA. the current through each resistance is sameB. the voltage through each resistance is sameC. neither current nor voltage through each resistance is sameD. both current and voltage through each resistance are same |
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Answer» Correct Answer - A In series combination current across its circuit components is always and in a parallel combination the volatage across the circuit components is constant |
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| 1825. |
Equivalent resistance (in ohm) of the given network is A. 28B. 18C. 26D. 25 |
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Answer» Correct Answer - B Between points A and B all resistance are combined in series `therefore R_("eq")=3Omega+4Omega+5Omega+6Omega=18Omega` |
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| 1826. |
In given circuit `R_(1)=5Omega,R_(2)=3Omega=7Omega` and supply voltage is 10V. Calculate power dissipated in `R_(1)` and `R_(3)`. |
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Answer» As voltage across `R_(1)` is 10V so power dissipated in `R_(1) P_(1)=(V^(2))/(R_(1))=(100)/(5)=20W` As `R_(2)` and `R_(3)` are connected in series so current in Power dissipated in `R_(3),P_(3)=I ^ (2)R_(3)=1xx7=7W R_(3)=(V)/(R_(2)+R_(3))=(10)/(3+7)=1A` |
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| 1827. |
Combine three resistors `5Omega, 4.5Omega and 3Omega` is such a way that the total resistance of this combination maximumA. `12.5Omega`B. `13.5Omega`C. `14.5Omega`D. `16.5Omega` |
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Answer» Correct Answer - A For maximum equivalent or total resistance of the resistors must be combined in series. `R_(eq)=R_(1)+R_(2)=R_(3)=5+4.5+3=12.5Omega` |
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| 1828. |
In the circuit shown here, cells A and B have emf 10 V each and the internal resistance is `5Omega` for A and `3Omega` for B. For what value of R will the potential difference across thhe cell A will be zero?A. zeroB. 1ohmC. 2ohmD. 3ohm |
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Answer» Correct Answer - C `i=(e_(1)-e_(2))/(R+r_(1)+r_(2)),v=e_(1)-ir_(1)=0_(7)` |
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| 1829. |
Two cells `e_(1) and e_(2)` connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance `3Omega` the cell is of emf 7V and internal resistance `7Omega`. The potential difference between the points A and B is A. 8.4VB. 5.6VC. 7.8VD. 6.6V |
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Answer» Correct Answer - A `I=(Deltaepsi)/(r_(1)+r_(2))=(9-7)/(3+7)=(2)/(10)=0.2A` Potential difference across cell `epsi_(1)` is `=9-0.2xx3=9-0.6=8.4V` Potential difference across `epsi_(2)`, `V_(AB")=epsi_(2)+0.2 r_(2)=7+0.2xx7=7+1,4=8.4V` |
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| 1830. |
The circuit in Fig. shows two cells connected in opposition to each other. Cell `E_(1)` is of emf `6V` and internal resistance `2Omega,` the cell `E_(2)` is of emf `4 V` sand internal resistance `8omega`. Find the potential difference between the points `A` and `B`. |
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Answer» Here, `E_(1) = 6 V, r_(1)= 2Omega, E_(2)=4 V, r_(2) = 8Omega` Refer Fig.`2(EP)`.12, current in circuit, `l=(6-4)/(2+8) = 1/5A` Potential difference across `A` and `B` is the terminal voltage of cell `E_(1) = E_(1) - Ir_(1) = 6-1/5xx2 = 5.6 V` |
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| 1831. |
A dry cell of emf `1.6V` and internal resistance of `0.10 Omega` is connected to a resistor of resistance `R omega`. If the current drawn the cell is `2A`, then (i) What is the voltage drop across R ? (ii) What is the rate of energy dissipation in the resistor ? |
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Answer» Here `epsilon = 1.5 V, r = 0.20 Omega, I = 1.5A` As `I = epsilon/(R + r) or R + r = epsilon/I` or `R = epsilon/I - r = 1.5/1.5 - 0.2 = 0.8 Omega` (a) Voltage drop across R, `V = IR = 1.5 xx 0.8 = 1.2 V` (b) Rate of energy dissipation inside the resistor `= I^(2)R = (1.5)^(2) xx 0.8 = 1.8W` |
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| 1832. |
The circuit shownin Fig. `7.33` contains a battery , a rheostat , and two identical lamps. What will happen to the brightness of the lamps if the resistance of the rheostat is increased? A. Less brighter BrighterB. Less brighter Loss brighterC. Brighter Less brighterD. No change Brighter |
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Answer» Correct Answer - A (a) Consider two extreme cases. (i) When the resistance of the rheostat is zero, the current through `Q` is zero since `Q` is short-circuited. The circuit is then essentially a battery in series with lamp `P`. (ii) When the resesitance of the rhostat is very large, amost no current flows throuh it. So, the currents through `P` and `Q` are almost equal. The current is essentially a battery in series with lamp `P` and `Q` |
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| 1833. |
The three resistances, each of value `5 Omega` are connected to the source of emf `epsilon` through ammeter A as shown in figure. If ammeter shows a reading of 2 A, calculate the power dissipated in the circuit. |
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Answer» Total resistance of the circuit is `R = ( 5 xx 5)/( 5 + 5) + 5 = 7.5 Omega` Current in the circuit, I = 2 A Power dissipated in the circuit, `P = I^(2)R = (2)^(2) xx 7.5 = 30.0W` |
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| 1834. |
In which one of the following arrangements of resistors does the ammeter `M`, the has a resistance of `2 Omega` give the largest reading when the same potentail difference is applied between point `P` and `Q`A. B. C. D. |
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Answer» Correct Answer - C (c ) Let `V_(PQ) = E` For `(A) : I = (E)/(4)` For `(B) : I = (E)/(2//3 + 2) = (E)/(8//3)` For `(C ) : I = (E)/(2)` For `(D) : I = (E)/(3)` |
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| 1835. |
The resistance of each of the three wires, The combination of resistors is connected to a source of emf `epsilon`. The ammeter shows a reading of `1A`. Calculate the power dissipated in the circuit |
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Answer» Correct Answer - `6 W` Total resistance of the circuit `R = (4 xx 4)/(4 + 4) + 4 = 6 Omega` current of the circuit `l = 1A` power dissipated in the circuit `P = l^(2) R = (1)^(2) xx 6 = 6 W` |
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| 1836. |
In meter bridge, the null points is found at a distance of 60.0 cm front end A. If now a resistance of `5 Omega` is connected in series with S, the null point occurs at 50 cm. Determine the values of R and S, |
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Answer» In first case, `R/S=60/(100-60) = 3/2` In second case, `R/S+5 = 50/50 = 1 or R = S+5` From (i), `S+5/S =3/2 or 2S+10 = 3S` or `S=10 Omega` and `R= 10+5 =15 Omega` |
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| 1837. |
A network of resistances is connected to a 16 V battery with internal resistance of `1 Omega`, as shown in Fig. 4.33. (a) Compute epuivalent resistance of the network, (b) obtain the current In in each resistor, and (c ) obtain the voltage drops `V_(AB), V_(BC) " and " V_(CD)`. |
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Answer» (a) The network is a simple series and parallel combination of resistors. First the two `4 Omega` resistors in parallel are equivalent to a resistor `= [(4 xx 4)//(4 + 4)] Omega = 2 Omega`. In the same way, the `12 Omega and 6 Omega` resistors in parallel are equivalent to a resistor of `[(12 xx 6)//(12 + 6)] Omega = 4 Omega`. The equivalent resistance R of the network is obtained by combining these resistors `(2 Omega and 4 Omega)` with `1 Omega` in series, that is, `R = 2 Omega + 4 Omega + 1 Omega = 7 Omega` (b) The total current `I` in the circuit is `I = (epsilon)/(R+r)=(16 V)/((7+1)Omega)=2A` Consider the resistors between A and B. If `I_(2)` is the current in one of the `4 Omega` resistors and `I_(2)` the current in the other `I_(1) xx 4 = I_(2)xx4` that is, `I_(1)=I_(2)`, which is otherwise obvious from the symmetry of the two arms. But `I_(1)+I_(2)=I=2A`. Thus, `I_(1)=I_(2)=1A` that is, current in each `4 Omega` resistor is 1A. current in `1 Omega` resistor between B and C would be 2A Now, consider the resistance between C and D. If `I_(3)` is the current in the `12 Omega` resistor, and `I_(4)` in the `6 Omega` resistor, `I_(3)xx12=I_(4)xx6` i.e., `I_(4)=2I_(3)` But, `I_(3)+I_(4)=I=2A` Thus, `I_(3)=((2)/(3))A,I_(4)=((4)/(3))A` that is, the current in the `12 Omega` resistor is `(2//3)A`, while the current in the `6 Omega` resistor is (4/3)A The voltage drop across AB is `V_(AB)=I_(1)xx4=1 A xx 4 Omega = 4V` This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is, `V_(AB) = 2 A xx 2 Omega = 4V` The voltage drop across BC is `V_(BC) = 2A xx 1 Omega = 2V` Finally, the voltage drop across CD is `V_(CD)=12 Omega xx I_(3)=12 Omega xx ((2)/(3))A=8V` This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D, that is, `V_(CD) = 2 A xx 4 Omega = 8 V` Note that the total voltage drop across AD is `4 V + 2 V + 8 V = 14 V`. Thus, the terminal voltage of the battery is 14 V, while its emf is 16 V. The loss of the voltage (= 2 V) is accounted for by the internal resistance `1 Omega` of the battery `[2 A xx 1 Omega= 2 V]`. |
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| 1838. |
The four arms of a Wheatstone bridge have the following resistance : `AB = 100 Omega, BC = 10 Omega, CD = 5 Omega`, and `DA = 60 Omega` A galvanometer of `15 Omega` resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. |
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Answer» Considering the mesh BADB, we have `100 I_(1) 15 I_(g) = 60 I_(2) = 0` or `20 I_(1) + 3I_(g) - 12 I_(2) = 0` Considering the mesh BCDB, we have `10 (I_(1) - I_(g)) - 15 I_(g) - 5 (I_(2) + I_(g)) = 0` `10 I_(1) - 30 I_(g) - 5I_(2) = 0` `2 I_(1) - 6 I_(g) - I_(2) = 0` considering the mesh ADCEA, `60 I_(2) + 5 (I_(2) + I_(g)) = 10` `65 I_(2) + 5 I_(g) = 10` `13 I_(2) + I_(g) = 2` Multiplying equation (2) by 0 `20 I_(1) + 60 I_(g) - 10 I_(2) = 0` Froms Equations (4) and (1) we have `63 I_(g) + 2 I_(2) = 0` `I_(2) = 31.5 I_(g)` Substituting tha value of `I_(2)` into Equation (3) we get. `13 (31. 5 I_(g)) + I_(g) = 2` `410.5 I_(g) = 2` `I_(g) = 4.87 mA`. |
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| 1839. |
In a meter bridge, the null points is found at a distance of 33.7 cm from A. If now a resistance of `12 Omega` is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S. |
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Answer» From the first balance point, we get `(R)/(S)=(33.7)/(66.3)` After S is connected in parallel with a resistance of `12 Omega`, the resistance across the gap changes from S to `S_(eq)`, where `S_(eq)=(12S)/(S+12)` and hence the new balance condition now gives `(51.9)/(48.1)=(R)/(S_(eq))=(R(S+12))/(12S)` Subsituting the value of `R//S` from Eq. (2.87), we get `(51.9)/(48.1)=(S+12)/(1).(33.7)/(66.3)` which gives `S= 13.5 Omega`. Using the value `R//S` above, we get `R = 6.86 Omega` |
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| 1840. |
The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: `AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega`. A galvanometer of 15W resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. |
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Answer» Considering the mesh BADB, we have `100 I_(1)+15 I_(g)-60 I_(2)=0` or `20 I_(1)+3I_(g)-12I_(2)=0` Considering the mesh BCDB, we have `10 (I_(1)-I_(g))-15 I_(g)-5(I_(2)+I_(g))=0` `10 I_(1)-30 I_(g)-5I_(2)=0` `2I_(1)-6I_(g)-I_(2)=0` Consideringn the mesh ADCEA, `60 I_(2)+5(I_(2)+I_(g))=10` `64 I_(2)+5I_(g)=10` `13 I_(2)+I_(g)=2` Multiplying Eq, (2.84b) by 10 `20 I_(1)-60 I_(g)-10 I_(2)=0` From Eqs. (2.84d) we have `63I_(g)-2I_(2)=0` `I_(2)=31.5 I_(g)` Substituting the value of `I_(2)` into Eq. [3.84 (c)], we get `13 (31.5 I_(g))+I_(g)=2` `410.5 I_(g)=2` `I_(g)=4.87 mA`. |
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| 1841. |
Two resistances are joined in parallel whose equivolent resistance is `3/5Omega`. One of the resistance wire is broken and the effective resistance becomes `3Omega`. The resistance (in ohms) of the wire that got broken wasA. `4/3`B. 2C. `6/5`D. `3/4` |
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Answer» Correct Answer - D Resistance of the wire `R_(P)=(R_(1)R_(2))/(R_(1)+R_(2))=(3)/(5)` and `R_(1)=3 Omega`, then `therefore (3xxR_(2))/(3+R_(2))=(3)/(5)rArr 15 R_(2)=9+3 R_(2)rArr 12 R_(2)=9` `therefore R_(2)=(3)/(4)Omega` |
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| 1842. |
Three resistances of magnitude 2, 3 and 5 ohm are connected in parallel to a battery of 10 volts and of negligible resistance. The potential difference across `3 Omega` resistance will beA. 2 voltsB. 3 voltsC. 5 voltsD. 10 volts |
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Answer» Correct Answer - D |
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| 1843. |
Two resistances are joined in parallel whose resultant is `6//8 ohm`. One of the resistance wire is broken and the effective resistance becomes `2 Omega` . Then the resistance in ohm of the wire that got broken wasA. `3//5`B. 2C. `6//5`D. 3 |
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Answer» Correct Answer - C |
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| 1844. |
Resistance `n`, each of `r ohm`, when connected in parallel give an equivalent resistance of `R ohm`. If these resistances were connected series, the combination would have a resistance in ohm, equal toA. `n^(2)R`B. `(R)/(n^(2))`C. `(R)/(n)`D. `nR` |
| Answer» Correct Answer - A | |
| 1845. |
When electric bulbs of same power, but different marked voltage are connected in series across the power line, their brightness will be :A. directly proportional to their marked voltages directly proportional to the squares of their marked voltagesB. inversely proportional to the squares of their marked voltagesC. directly proportional to the squares of their marked voltagesD. inversely proportional to the squares of their marked voltages. |
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Answer» For a bulb, power `(P)=(V^(2))/(R)` `therefore P=(V_(1)^(2))/(R_(1))=(V_(2)^(2))/(R_(2))` `therefore R_(1)=(V_(1)^(2))/(P)` and `R_(2)=(V_(2)^(2))/(P)` The seres current through each is same. Heat `prop I^(2) Rt`. or `(H_(1))/(H_(2))=(R_(1))/(R_(2))` or `(H_(1))/(H_(2))=(V_(1)^(2))/(V_(2)^(2))` or `H propV^(2)` or Brightness `porp("voltage")^(2)`. |
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| 1846. |
When three identical bulbs of `60 W, 200 V` rating are connected in series to a `200 V` supply, the power drawn by them will beA. 180 wattB. 10wattC. 20 wattD. 60watt |
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Answer» If rated voltage=supply voltage then in series combination of bulbs `(1)/(P_(eq))=(1)/(P_(1))+(1)/(P_(2))+(1)/(P_(3))+`............. For this question `(1)/(P_(eq))=(1)/(60)+(1)/(60)+(1)/(60)=(3)/(60)=(1)/(20)rArrP_(eq)=20` watt |
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| 1847. |
When three identical bulbs of `60 W, 200 V` rating are connected in series to a `200 V` supply, the power drawn by them will beA. `60 W`B. `180 W`C. `10 W`D. `20 W` |
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Answer» Correct Answer - D (d) Let `R_(1), R_(2)` and `R_(3)` are the resistance of three bulbs respectively. In series order, `R = R_(1) + R_(2) + R_(3)` but `R = V_(2)//P` and supply voltage in series order is the same as the rated voltage. `:. (V^(2))/(P) = (V^(2))/(P_(1)) + (V^(2))/(P_(2)) + (V^(2))/(P_(3))` or `(1)/(P) = (1)/(60) + (1)/(60) + (1)/(60)` or `P = (60)/(3) = W` |
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| 1848. |
Two batteries, one of emf 18 volts and internal resistance `2 Omega` and the other fo emf 12 volts and internal resistance `1 Omega`, are connected as shown. The voltmeter `V` will record a reading of A. 15 voltB. 30 voltC. 14 voltD. 18 volt |
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Answer» Correct Answer - C (c ) Reading of voltmeter `= E_(eq) = (E_(1) r_(2) + E_(2) r_(2))/(r_(1) + r_(2)) = (18 xx 1 + 12 xx 2)/(1 + 2) = 14 V` |
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| 1849. |
Shown in the figure below is a meter- bridge set up will null deflection in the galvanometer. The value of the unknown resistor R isA. `13.75 Omega`B. `220 Omega`C. `110 Omega`D. `55 Omega` |
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Answer» Correct Answer - B (b) According to the condition of balancing `55/20 = R/80 rArr R = 220 Omega`. |
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| 1850. |
Shown in the figure below is a meter- bridge set up will null deflection in the galvanometer. The value of the unknown resistor R isA. `13.75Omega`B. `220Omega`C. `110Omega`D. `55Omega` |
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Answer» Correct Answer - B `i=(E)/(R)` |
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