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 __________.
คำตอบ
5
คำอธิบาย
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 VAB by considering the upper path
For the upper path,
$${V_A} - 6 + 1 \times 1 = {V_B}$$
$${V_A} - {V_B} = 5V$$