Question

A battery of $ 12\,V $ and an internal resistance of $ 0.80 $ is connected to 3 resistors as shown in the fig. The current ( in the circuit is

• A. 2.3 A
• B. 4.0 A
• C. 2.0 A
• D. 1.33 A

Answer 1

Answer: (C)

Using the relation for current
$i=\frac{E}{R +r}$ ...(1)
But the equivalent resistance in the upper arm is given by as
$R_{\text {equivalent }}=4\, \Omega+\frac{2 \times 3}{2+3} \Omega$
$R_{q}=4+\frac{6}{5}=\frac{26}{5}=5.2$
Now equation (1) we get
$i=\frac{12}{5.2+0.8}=\frac{12}{6}=2\, amp$