Question

Consider the circuit shown below. The current in the $4\Omega$ resistor is

• A. $0.25\text{A}$
• B. $0.50\text{A}$
• C. $0.75\text{A}$
• D. $1.00\text{A}$

Answer 1

Answer: (A)

Given circuit below

Resistance between $B$ and $C$ on the right-hand side of the circuit
$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$
because $1\Omega +4\Omega +5\Omega =10\Omega$ are in parallel with $10\Omega$ .
Resistance between $A$ and $D$ on the right-hand side of the circuit.
$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\Omega$
because $5\Omega$ (equivalent resistance of $1\Omega ,4\Omega ,5\Omega$ and $10\Omega$ ) is in series with $3\Omega$ and $2\Omega$ .
Hence, resistance between $A$ and $D$ will be $5\Omega$
Equivalent resistance of the circuit,
$R=5+2+5=12\Omega$
Current drawn from the battery,
$I=\frac{12V}{12\Omega }=1\text{A}$
At the junction $\text{A}$ , the current of $1\text{A}$ is divided equally between the $10\Omega$ resistance and the remaining circuit of the $10\Omega$ resistance.
At the junction $B$ , the current of $0.5\text{A}$ is divided equally between the $10\Omega$ resistor and remaining circuit of resistance $10\Omega$ .
$\therefore$ Current through the $4\Omega$ resistor $=0.25$ .