Question

In the following circuit diagram, $E=4V,r=1\Omega$ and $R=45\Omega$, then reading in the ammeter A will be

• A. $1A$
• B. $\frac{1}{2}A$
• C. $\frac{1}{8}A$
• D. $\frac{1}{4}A$

Answer 1

Answer: (D)

$\frac{1}{4}A$

The circuit can be redrawn as shown in figure.

The equivalent resistance in the circuit is

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}$

$\frac{1}{R_{eq}}=\frac{1}{45}+\frac{1}{45}+\frac{1}{45}$

$\frac{1}{R_{eq}}=\frac{3}{45}$

$R_{eq}=15\Omega$

Now the current in the circuit is

$I=\frac{E}{r+R_{eq}}$

$I=\frac{4}{1+15}$

$I=\frac{4}{16}$

$I=\frac{1}{4}A$