Question

The current (in A) flowing through $1\Omega$ resistor in the following circuit is

• A. $0.40A$
• B. $0.25A$
• C. $0.20A$
• D. $0.5A$

Answer 1

The correct option is C

$0.20A$

Step 1:Given data and figure

Resistors $R=1$ohm

Step 2: Use the formula

Now used the formula,

$I_{total}=\frac{V}{R_{total}}$

Where $V$ is the potential difference.

The potential difference in the top most branch is $V=1V$

$R_{1}and{R}_2$are one Ohm each.

Now calculate the total resistance of the branch.

Use the formula for the effective resistance in series.

$R_{total}=R_p+2\Omega .............\left(1\right)$

By using the formula for resistance in parallel.

$R_p=\left(\frac{R_1\times R_2}{R_1+R_2}\right)$

Step 3: Solution

Now substituting the values of $R_{1}and{R}_2$

$$\begin{gathered} \Rightarrow R_p=\left(\frac{1\times 1}{1+1}\right) \\ \Rightarrow R_p=0.5\Omega \end{gathered}$$

Substitute the value in equation (1)

$$\begin{gathered} R_{total}=0.5+2 \\ {R}_{total}=2.5\Omega \end{gathered}$$

According the Ohms law,

$$\begin{gathered} \Rightarrow I_{total}=\frac{V}{R_{total}} \\ \Rightarrow I_{total}=\frac{1}{2.5} \\ \Rightarrow I_{total}=0.4A \end{gathered}$$

Note that the potential differences in the two branches are the same and the current divides into $I1andI2$step

Step 3: Use the formula

$\Rightarrow V=I_1{R}_1=I_2{R}_2$

The total current,

$\begin{array}{rcl}& \Rightarrow & I=I_1+I_2\\ & \Rightarrow & I_1{R}_1=I_2{R}_2\\ & \Rightarrow & I_1{R}_1=\left(I-I_1\right)R_2\end{array}$

Step 3: Calculate $I_1$

Therefore,

$\Rightarrow I_1(R_1+R_2)=I_2{R}_2$

Now,

$$\begin{gathered} \Rightarrow I_1=\frac{I_2{R}_2}{\left(R_1+R_2\right)} \\ \Rightarrow I_1=\frac{0.4\times 1}{\left(1+1\right)} \\ \Rightarrow I_1=0.2A \end{gathered}$$

Hence, the correct option is (C).