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

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

Tardigrade Admin · Accepted Answer

Using the relation for current i=(E/R +r) ...(1) But the equivalent resistance in the upper arm is given by as R text equivalent =4 Ω+(2 × 3/2+3) Ω Rq=4+(6/5)=(26/5)=5.2 Now equation (1) we get i=(12/5.2+0.8)=(12/6)=2 amp

Q. 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

2.3 A

4.0 A

2.0 A

1.33 A

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

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