Question

constant volume of 50v is maintained between points a & b of circuit shown in the figure current rough branch cd of circuit is: 1.5a 2) 2a 3) 2.5a 4) 3.0a

Explanation:

Step1: Calculate equivalent resistance of upper - left and lower - left resistors

The 1Ω and 3Ω resistors are in parallel. The equivalent resistance $R_1$ of two resistors $R_a = 1\Omega$ and $R_b=3\Omega$ in parallel is given by $\frac{1}{R_1}=\frac{1}{1}+\frac{1}{3}=\frac{3 + 1}{3}=\frac{4}{3}$, so $R_1=\frac{3}{4}\Omega$.

Step2: Calculate equivalent resistance of upper - right and lower - right resistors

The 2Ω and 4Ω resistors are in parallel. The equivalent resistance $R_2$ of two resistors $R_c = 2\Omega$ and $R_d = 4\Omega$ in parallel is given by $\frac{1}{R_2}=\frac{1}{2}+\frac{1}{4}=\frac{2+1}{4}=\frac{3}{4}$, so $R_2=\frac{4}{3}\Omega$.

Step3: Calculate total equivalent resistance of the circuit

$R_1$ and $R_2$ are in series. The total equivalent resistance $R = R_1+R_2=\frac{3}{4}+\frac{4}{3}=\frac{9 + 16}{12}=\frac{25}{12}\Omega$.

Step4: Calculate total current in the circuit

Using Ohm's law $I=\frac{V}{R}$, with $V = 50V$ and $R=\frac{25}{12}\Omega$, we have $I=\frac{50}{\frac{25}{12}}=24A$.

Step5: Use current - division rule

The voltage across $R_1$ and $R_2$ is divided in proportion to their resistances. Since $R_1$ and $R_2$ are in series, the voltage across $R_2$ is $V_2=\frac{R_2}{R_1 + R_2}\times V$. Since $R_1=R_2=\frac{4}{3}\Omega$, $V_2 = 25V$.
The current through the 2 - 4Ω parallel combination is $I_{2 - 4}=\frac{V_2}{R_2}=\frac{25}{\frac{4}{3}}=\frac{75}{4}A$.
Using current - division rule for the 2Ω and 4Ω resistors in parallel, the current through the 2Ω resistor (branch CD) is $I_{CD}=\frac{4}{2 + 4}\times\frac{75}{4}= \frac{75}{6}= 2.5A$.

Answer:

1. 2.5A