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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.
| 101. |
The laws of elctromagnetic induction have bee used in the construction of aA. galvanometerB. voltmeterC. electric motorD. electric generator |
| Answer» Correct Answer - D | |
| 102. |
A closedc coil with a resitance `R` is placed in a magnetic field. The flux linked with the coil is `phi`. If the magnetic field is suddenly reversed in direction, the charge that flows through the coil will beA. `phi//2 R`B. `phi//R`C. `2 phi//R`D. zero |
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Answer» Correct Answer - C `q = (d phi)/(R )` |
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| 103. |
A square loop of side `22 cm` is changed to a circle in time `0.4 sec` with its plane normal to a magnetic field `0.2 T`. The emf induced isA. `+6.6 m v`B. `-6.6 m v`C. `+13.2 m v`D. `-13.2 m v` |
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Answer» Correct Answer - B `e = -B(Delta A)/(Delta t)` |
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| 104. |
A coil has `1,000` turns and `500 cm^(2)` as its area. The plane of the coil is placed at right angles to a magnetic induction field of `2 xx 10^(-5) web//m^(2)`. The coil is rotated through `180^(@)` in `0.2` second. The average emf induced in the coil, in milli volts, is :A. 5B. 10C. 15D. 20 |
| Answer» Correct Answer - B | |
| 105. |
a coil of `1200` turns and mean area of `500 cm^(2)` is held perpendicular to a uniform magnetic field of induction `4 xx 10^(-4) T`. The resistance of the coil is `20` ohms. When the coil is rotated through `180^(@)` in the magnetic field in `0.1` seconds the average electric current (in `mA`) induced is :A. `12`B. `24`C. `36`D. `48` |
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Answer» Correct Answer - B `E = (2NBA)/(t)`, `i = (E)/(R )` |
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| 106. |
A coil has `1,000` turns and `500 cm^(2)` as its area. The plane of the coil is placed at right angles to a magnetic induction field of `2 xx 10^(-5) web//m^(2)`. The coil is rotated through `180^(@)` in `0.2` second. The average emf induced in the coil, in milli volts, is :A. `5`B. `10`C. `15`D. `20` |
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Answer» Correct Answer - B `e = (2NBA)/(t)` |
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| 107. |
Two coils `A` and `B` having turns `300` and `600` respectively are placed near each other, on passing a current of `3.0` ampere in `A`, the flux linked with `A` is `1.2xx10^(-4)` and with `B` it is `9.0xx10^(-5)` weber. The mutual inuctance of the system isA. `2 xx 10^(-5)` henryB. `3 xx 10^(-5)` henryC. `4xx 10^(-5)` henryD. `6 xx 10^(-5)` henry |
| Answer» Correct Answer - B | |
| 108. |
A field of strenght `5 xx 10^(4)//pi` ampere turns`//`meter acts at right angles to the coil of `50` turns of area`10^(-2) m^(2)`. The coil is removed from the field in `0.1` second. Then the induced `e.m.f` in the coil isA. `0.1 V`B. `0.2 V`C. `1.96 V`D. `0.98 V` |
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Answer» Correct Answer - A `B = mu_(0) H = (mu_(0) xx 5 xx 10^(4))/(pi)` `e = (NBA)/("time") = (50 xx 2 xx 10^(-2) xx 10^(-2))/(0.1) = 0.11` |
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| 109. |
A rod `PQ` is connected to the capacitor plates. The rod is placed in a magnetic field `(B)` directed downwards perpendicular to the plane of the paper. If the rod is pulled out of magnetic field with velocity `vec(v)` as shown in Figure. A. Plate `M` will be positively chargedB. Plate `N` will be positively chargedC. Both plates will be similarly chargedD. no charge will be collected on plates. |
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Answer» Correct Answer - A Consider the force on n electron in `PQ`. This electron experiences a force towards `Q`. Free electron in `PQ` tend to move towards `N`. So `M` will be positively charged. |
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| 110. |
The inductance between A and D is A. `3.66 H`B. `9 H`C. `0.66 H`D. `1 H` |
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Answer» Correct Answer - D The given arrangement is a parallel combination of inductances. Therefore, `(1)/(L)=(1)/(3)+(1)/(3)+(1)/(3) = (3)/(3) = 1` `rArr L = 1 H` |
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| 111. |
The equivalent quantity of mass in electricity isA. ChargeB. PotentialC. InductanceD. Inductance |
| Answer» Correct Answer - C | |
| 112. |
The momentum in mechanics is expressed as `m xx v` . The analogous expression in electricity isA. `I xx Q`B. `I xx V`C. `L xx I`D. `L xx Q` |
| Answer» Correct Answer - C | |
| 113. |
An inductor of 5 H inductance carries a steady current of 2 A. How can a 50 V self induced e.m.f. be made to appear in the inductor ? |
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Answer» `L = 5H, |e| = 50V`, Let us produce the required emf by reducing current to zero Now, `|e| = L(dI)/(dt)` or `dt = (LdI)/(|e|) = (5 xx 2)/(50) s` `(10)/(50)s = (1)/(8)s = 0.2 s` So, the desireed emf can be produced by reducing the given current to zero in `0.2` second. |
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| 114. |
Assertion : An artificial satellite with a metal surface is moving above the earth in a circular orbit. A current will be induced in satellite if the plane of the orbit is inclined to the plane of the equator. Reason : The current will be induced only when the speed of satellite is more than `8 km//sec`.A. If both assertion and reason are t rue and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false |
| Answer» Correct Answer - C | |
| 115. |
A square coil of `10^(-2) m^(2)` area is placed perpenducular to a uniform magnetic field of `10^(3) Wb//m^(2)`. What is magnetic flux through the coil?A. 10 weberB. `10^(-5)` weberC. `10^(5)` weberD. 100 weber |
| Answer» Correct Answer - A | |
| 116. |
Work of electric motor isA. To convert ac into dcB. To convert dc into acC. Both (a) and (b)D. To convert ac into mechanical work |
| Answer» Correct Answer - D | |
| 117. |
In electromagnetic induction, the induced charge in a coil is independent ofA. Change in the fluxB. TimeC. Resistance in the circuitD. None of the above |
| Answer» Correct Answer - B | |
| 118. |
In an induction coil with resistance, the induced emf will be maximum whenA. The switch is put on due to high resistanceB. The switch is put off due to high resistanceC. The switch is put on due to low resistanceD. The switch is put off due to low resistance |
| Answer» Correct Answer - B | |
| 119. |
Quantity that remains unchanged in a transformer isA. VoltageB. CurrentC. FrequencyD. None of the above |
| Answer» Correct Answer - C | |
| 120. |
When a bar magnet falls through a long hollow metal cylinder fixed with its axis vertical, the final acceleration of the magnet isA. Equal to zeroB. Less than gC. Equal to gD. Equal to g in to beginning and then more than g |
| Answer» Correct Answer - A | |
| 121. |
The magnetic flux linked with a circuit of resistance `100 ohm` increase from `10` to `60` webers. The amount of induced charge that flows in the circuit is (in coulomb)`A. `0.5`B. 5C. 50D. 100 |
| Answer» Correct Answer - A | |
| 122. |
A conducting wire is moving towards right in a magnetic field `B`. The direction of induced current in the wire is shown in the figure. The direction of magenetic field will be A. In the plane of paper pointing towards rightB. In the plane of paper pointing towards leftC. Perpendicular to the plane of paper and down-wardsD. Perpendicular to the plane of paper and upwards |
| Answer» Correct Answer - C | |
| 123. |
Shows a copper rod moving with velocity `v` parallel to a long straight wire carrying current `= 100 A`. Calculate the induced emf in the rod, where `v = 5 m S^(-1)`, `a = 1 cm` , `b = 100cm`. A. `0.23 mV`B. `0.46 mV`C. `0.16 mV`D. `0.32m V` |
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Answer» Correct Answer - B Let there be an an element `dx` of rod at a distance `x` from the wire. emf developed in the element, `dE = Bdxv` `:. dE = ((mu_(0))/(4pi)(2I)/(x))dx v` `:. E = (mu_(0)Iv)/(2pi)int_(a)^(b)(dx)/(x) = (mu_(0)Iv)/(2pi)"log"_(e)(b)/(a)` `:.E = (4pi xx 10^(-7) xx 100 xx 5)/(2pi) "log"_(e) (100)/(1)` `= 4.6 xx 10^(-4) V = 0.46 mV` |
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| 124. |
In which of the following cases the emf is induced due to time varying magnetic feld (induced field emf)? Case I A magnet is moving along the axis of a conducting coil Case II A loop having varying area (due to moving jumper) is placed in a magnetic field case III The resistance of the coil is changing, which is connected to an ideal battery. case IV a current carrying wire is approaching a conducting ring.A. `I,II` and `III` onlyB. `I,III` and `IV` onlyC. `I,II` and `IV` onlyD. All the four |
| Answer» Correct Answer - B | |
| 125. |
A pure inductor `L`, a capactior `C` and a resistance `R` are connected across a battery of emf `E` and internal resistance `r` as shows in Fig. Switch `S_(W)` is closed at `t = 0`, select the correct altermative (S). A. current through resistance `R` is zero all the timeB. current through resistance `R` is zero at `t = 0` and `t rarr oo`C. maximum charge stored in the capacitor is `CE`D. maximum energy stored in the indcutor is equal to the maximum energy stored in the capacitor |
| Answer» Correct Answer - B | |
| 126. |
The current carrying wire and the rod AB are in the same plane. The rod moves parallel to the wire with a velocity v . Which one of the following statements is true about induced emf in the rod A. End A will be at lower potential with respect to BB. A and B will be at the same potentialC. There will be no induced e.m.f. in the rodD. Potential at A will be higher than that at B |
| Answer» Correct Answer - D | |
| 127. |
A conducting rod length `1m` moves with a speed of `2m//s` parallel to a straight long wire carrying `4 A` current. Axis of rod is kept perpendicular to the wire with its near end at a distance of `1m` from the wire. Magnitude of induced emf in rod is `N xx 10^(-7) log 16 "volt"`. Find the value of `N`. |
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Answer» Correct Answer - D Field at distance `x` from the wire is `B = (mu_(0)i)/(2pi x)` and `phi_(0) = int B.dA` `:. phi_(B) = int_(l)^(2l)(mu_(0)i vt)/(2pi) ((dx)/x)=(mu_(0)i vt)/(2pi)"log"2` Hence induced emf is `E = (d phi_(B))/(dt) = (mu_(0)i v)/(2pi)"log" 2 = 16 xx 10^(-7) log2` `:. N = 4` |
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| 128. |
In the circuit shown in Fig. A conducting wire HE is moved with a constant speed v towards left. The complete circuit is placed in a uniform magnetic field `vec(B)` perpendicular to the plane of circuit inwards. The current in HKDE is A. clockwiseB. anticlockwiseC. alternatingD. zero |
| Answer» Correct Answer - D | |
| 129. |
A flexible conducting wire in the form of a circle is kept in a uniform magnetic field with its plane normal to the field. Radius of that circle changes with time as shown. Then which of the following graphs represents the variation of induction emf with time `R = R_(0), t lt t_(0) , R = R_(0) + t, t_(0) lt t lt 2t_(0)`: A. B. C. D. |
| Answer» Correct Answer - C | |
| 130. |
A conducting loop of radius `R` is present in a uniform magnetic field `B` perpendicular to the plane of ring. If radius `R` varies as a function of time `t` as `R = R_(0)+t^(2)`. The emf induced in the loop is A. `2piBt(R_(0)+t^(2))` ClockwiseB. `2piBt(R_(0)+t^(2))` AnticlockwiseC. `4piBt(R_(0)+t^(2))` AnticlockwiseD. `4 pi B t(R_(0)+t^(2))` Clockwise |
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Answer» Correct Answer - C `e = -(d phi)/(dt) = pi B(d)/(dt)(R^(2))` |
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| 131. |
A coil having an area `A_(0)` is placed in a magnetic field which changes from `B_(0)` to `4B_(0)` in a time interval `t`. The e.m.f. induced in the coil will beA. `(3A_(0)B_(0))/(t)`B. `(4A_(0)B_(0))/(t)`C. `(3B_(0))/(A_(0)t)`D. `(4B_(0))/(A_(0) t)` |
| Answer» Correct Answer - A | |
| 132. |
Two capacitors of capacitance `C` and `3C` are charged to potential difference `V_(0)` and `2V_(0)` respectively, and connected to an inductor of inductance `L` as shown in figure. Initially the current in the inductor is zero. Now, the switch `S` is closed. Potential difference across capacitor of capacitance `C` when the current in the cirucit is maximum isA. `(V_(0))/(4)`B. `(3V_(0))/(4)`C. `(5V_(0))/(4)`D. `(7V_(0))/(4)` |
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Answer» Correct Answer - C When current is maximum `(dI)/(dt) = 0` e.m.f of across `L = 0` so potential difference across the capacitor will be same. From the law of conservation of charge on plate `2` and `3`. `3CV + CV = 6CV_(0) - CV_(0) rArr V = (5)/(4)V_(0)` |
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| 133. |
Two capacitors of capacitance `C` and `3C` are charged to potential difference `V_(0)` and `2V_(0)` respectively, and connected to an inductor of inductance `L` as shown in figure. Initially the current in the inductor is zero. Now, the switch `S` is closed. The maximum current in the indcutor is :A. `(3V_(0))/(2) sqrt((3C)/(L))`B. `V_(0) sqrt((3C)/(L))`C. `2V_(0) sqrt((3C)/L)`D. `V_(0) sqrt((C )/(L))` |
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Answer» Correct Answer - A Loss in energy of capacitor = Energy store in inductor `rArr (1)/(2)VC_(0)^(2) + (1)/(2)3C(2V_(0))^(2)-(1)/(2) xx 4 CV^(2) = (1)/(2)LI` `rArr I = (3)/(2)V_(0)sqrt((3V_(0))/(L))` |
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| 134. |
Two capacitors of capacitance `C` and `3C` are charged to potential difference `V_(0)` and `2V_(0)` respectively, and connected to an inductor of inductance `L` as shown in figure. Initially the current in the inductor is zero. Now, the switch `S` is closed. Potential difference across capacitor of capacitance `3C` when the current in the cirucit is maximum is :A. `(V_(0))/(4)`B. `(V_(0))/(4)`C. `(5V_(0))/(4)`D. `(7V_(0))/(4)` |
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Answer» Correct Answer - C When current is maximum `(dI)/(dt) = 0` e.m.f of across `L = 0` so potential difference across the capacitor will be same. From the law of conservation of charge on plate `2` and `3`. `3CV + CV = 6CV_(0) - CV_(0) rArr V = (5)/(4)V_(0)` |
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| 135. |
The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases asA. `1` and `A` increasesB. `1` decreases and `A` increasesC. `1` increases and `A` decreasesD. both `1` and `A` decreases |
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Answer» Correct Answer - B The self-inductance of a long solenoid of cross-sectinal area `A` and length `1`, having `n` turns per unit length, filled the inside of `s` the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) is given by `L = mu_(r ) mu_(0)n^(2)Al` where, `" "n = N//1` |
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| 136. |
The self inductance of a solenoid of length L, area of cross-section A and having N turns is-A. `(mu_(0)N^(2)A)/(L)`B. `(mu_(0)NA)/(L)`C. `mu_(0)N^(2)LA`D. `mu_(0)NAL` |
| Answer» Correct Answer - A | |
| 137. |
When the number of turns and the length of the solenoid are doubled keeping the area of cross-section same, the inductanceA. Remains the sameB. Is halvedC. Is doubledD. Becomes four times |
| Answer» Correct Answer - C | |
| 138. |
The time constant of the circuit shown is A. `RC`B. `2RC`C. `3RC`D. `4RC` |
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Answer» Correct Answer - B `R^(1) = (R )/(2)+R = (3R)/(2) , i = RC = (3R)/(2) xx (4C)/(3) = 2RC` |
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| 139. |
In the circuit shown below, the key K is closed at t =0. The current through the battery is A. `(VR_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) "at t = 0"` and `(V)/(R_(2)) "at t" = oo`B. `(V)/(R_2) "at t = 0"` and `(V(R_(1)+R_(2)))/(R_(1)R_(2)) "at t" = oo`C. `(V)/(R_2) "at t = 0"` and `(VR_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) "at t" = oo`D. `(V(R_(1)+R_2))/(R_(1)R_(2)) "at t = 0"` and `(V)/(R_(2)) "at t" = oo` |
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Answer» Correct Answer - B At `t = 0`, inductor behaves like an infinite resistance so at `t = 0, i = (V)/(R_(2))` and at `t = oo`, inductor behaves line an a conducting wire `i = (V)/(R_("eq")) = (V(R_(1)+R_(2)))/(R_(1)R_(2))` |
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| 140. |
Assertion: Only a charge in magnetic flux will maintain an induced current in the coil. Reason: The presence of large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.A. Both `A` and `R` are true and `R` is the correct explanation of `A`B. Both `A` and `R` are true and `R` is not the correct explanation of `A`C. `A` is true but `R` is falseD. `A` is false but `R` is true. |
| Answer» Correct Answer - C | |
| 141. |
A chock coil has an inductance of `4H` and a resistance of `2 Omega`. It is connected to a battery of `12 V` and negaligible internal resistance. The time taken for the current to become `3.78 A` is nearlyA. `8 s`B. `.^(1)//_(2)`C. `2 s`D. `4 s` |
| Answer» `i = i_(0)[1-e^(-(R )/(L)t)]` | |
| 142. |
A coil resistance `20 Omega` and inductance `5 H` is connected with a `100 V` battery. Energy stored in the coil will beA. `41.5 J`B. `62.50 J`C. `125 J`D. `250 J` |
| Answer» Correct Answer - B | |
| 143. |
In L - R circuit, for the case of increasing current, the magnitude of current can be calculated by using the formulaA. `I = I_(0)e^(-Rr//L)`B. `I = I_(0)(I-e^(-Rt//L))`C. `I = I_(0)(I-e^(Rt//L))`D. `I = I_(0)e^(Rt//L)` |
| Answer» Correct Answer - B | |
| 144. |
In the figure magnetic energy stored in the coil is A. ZeroB. InfiniteC. 25 joulesD. None of the above |
| Answer» Correct Answer - C | |
| 145. |
When the current changes from `+2A` to `-2A` in `0.05s`, and emf of `8B` is induced in a coil. The coefficient of self-induction of the coil isA. `0.1H`B. `0.2 H`C. `0.4H`D. `0.8 H` |
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Answer» Correct Answer - A `8 = L(4)/(0.05)` `rArr L = 2 xx 0.05H = 0.1 H` |
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| 146. |
The magnetic induction in the region between the pole faces of an electromagnet is 0.7 weber/m 2. The induced e.m.f. in a straight conductor 10 cm long, perpendicular to B and moving perpendicular both to magnetic induction and its own length with a velocity 2 m/sec isA. `0.08 V`B. `0.14 V`C. `0.35 V`D. `0.07 V` |
| Answer» Correct Answer - B | |
| 147. |
A coil of area `10cm^2` and 10 turns is in magnetic field directed perpendicular to the plane and changing at a rate of `10^8 gauss//s`. The resistance of coil is `20Omega`. The current in the coil will beA. 5 ampB. `0.5 amp`C. `0.05 amp`D. `5 xx 10^(8) amp` |
| Answer» Correct Answer - A | |
| 148. |
A coil of area `10cm^2` and 10 turns is in magnetic field directed perpendicular to the plane and changing at a rate of `10^8 gauss//s`. The resistance of coil is `20Omega`. The current in the coil will beA. `0.5 A`B. `5A`C. `50 A`D. `5 xx 10^(8)A` |
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Answer» Correct Answer - B `i = (1)/(R )NA.(dB)/(dt) = (1)/(20) xx 10 xx 10^(-3) xx 10^(4)` |
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| 149. |
The resistance in the following circuit is increase at a particle instant. At this instant the value of resistanc eis `10Omega`. The current in the circuit will be now A. `i = 0.5 A`B. `i gt 0.5 A`C. `i lt 0.5 A`D. `i = 0` |
| Answer» Correct Answer - B | |
| 150. |
A coil of inductance `1 H` and resistance `10Omega` is connected to a resistanceless battery of emf `50 V` at time `t=0`. Calculate the ratio of rthe rate which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at `t=0.1s`.A. `e`B. `(1)/(e-1)`C. `1-(1)/(e)`D. `(1)/(e)` |
| Answer» `((dU)/(dt))/(p) = (1)/(2)(L(2i)(di)/(dt))/(Vi) = (L)/(V)i_(0)(R )/(L)e^(-(R)/(L)t) = (R )/(V)((V)/(R ))e^(-1) = (1)/(e)` | |