JEE Advance - Physics (2011 - Paper 2 Offline - No. 17)

Two batteries of different emfs and different internal resistance are connected as shown. The voltage across AB in volts is __________.

IIT-JEE 2011 Paper 2 Offline Physics - Current Electricity Question 12 English

Risposta

IIT-JEE 2011 Paper 2 Offline Physics - Current Electricity Question 12 English Explanation

Applying Kirchhoff's second law for closed loop CDEFC we get

$$ - 3 - 2I - I + 6 = 0$$

$$I = {{6 - 3} \over 3} = 1A$$

For the lower path

$${V_A} - 3 - 2 \times 1 = {V_B}$$

$$\therefore$$ $${V_A} - {V_B} = 5V$$

We can also find the V_AB by considering the upper path

For the upper path,

$${V_A} - 6 + 1 \times 1 = {V_B}$$

$${V_A} - {V_B} = 5V$$