Saved Bookmarks
| 1. |
A battery of internal resistance 3 ohm is connected to 20 ohm resistor and potential difference across the resistor is 10v if another resistor of 30 is connected in series with first resistor and battery is again connected to the combination |
|
Answer» complete the question properly .The question should have been like this :"A battery of internal resistance 3 ohm is connected to 20 ohm resistor and potential difference across the resistor is 10v if another resistor of 30 is connected in series with first resistor and battery is again connected to the COMBINATION . Calculate the EMF and potential difference across combination "Let the current flowing in the circuit be i and E be the emf of battery By applying Kirchhoff's Voltage law : E-20i-3i=0 E=23i But according to the question : Voltage across 20 ohms is 10V by ohms' law V=IR 10 = i x 20 i= 10/20 = 1/2 A= 0.5 A So emf of battery is E= 23 x i=23x 0.5= 11.5V now if 30 ohms resistance is connected then : Total resistance = 20+30+3=53 OHMS CURRENT : I=E/R=11.5/53=0.216A potential drop across combination of RESISTORS : v=i(20+30)\v= 50X 0.216=10.48V |
|