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Current Electricity

Question
CBSEENPH12047917
Wired Faculty App

Two batteries of emf 3V and 6 V with internal resistances 2 Ω and 4 Ω are connected in a circuit with resistance of 10 Ω as shown in figure. The current and potential difference between the points P and Q are

  • 316 A and 815 V

  • 163 A and 158 V

  • 316 A and 8 V

  • 316 A and 158 V

Solution

D.

316 A and 158 V

Kirchoff's voltage law states that for a closed loop series path, the algebraic sum of all voltages around any closed loop in a circuit is equal to zero.

Applying Kirchhoff's voltage law in the given loop 

   -- 4I + 6 - 3 - 2I - 10I  = 0

                    -16 I = -3

                      I = 316 A

∵  Potential difference across PQ

          vPQ = 316 × 10 V

             vPQ= 158

Some More Questions From Current Electricity Chapter

What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current A

Voltage V

Current A

Voltage V

0.2

0.4

0.6

0.8

1.0

2.0

3.94

7.87

11.8

15.7

19.7

39.4

3.0

4.0

5.0

6.0

7.0

8.0

59.2

78.8

98.6

118.5

138.2

158.0

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