Thank you for reporting, we will resolve it shortly
Q.
Consider the circuit shown below. The current in the $4 \, \Omega $ resistor is
NTA AbhyasNTA Abhyas 2020Current Electricity
Solution:
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,
$I=\frac{12 \, V}{12 \, \Omega }=1 \, \text{A}$
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 $=0.25\text{}$ .
