Question

Figure shows a two branched parallel circuit with $R_A = 10 \Omega$, $L = \frac{\sqrt{3}}{10} H$, $R_B = 20 \Omega$ and $C = \frac{\sqrt{3}}{2} mH$. Current in $L-R_A$ is $I_1$ and in $C-R_B$ is $I_2$ and main current is I

• A. Phase difference between $I_1$ and $I_2$ is $90^\circ$
• B. At some instant current in $L-R_A$ is 10 A. At the same instant current in $C-R_B$ branch will be $5\sqrt{3}$ A
• C. At some instant $I_1$ is $10\sqrt{2}$ A then at this instant I will be $10\sqrt{2}$ A
• D. Power dissipated in the circuit is 2121.3 W

Answer 1

Answer: (A), (B), (C)

Phase difference between $I_1$ and $I_2$ is $90^\circ$, At some instant current in $L-R_A$ is 10 A. At the same instant current in $C-R_B$ branch will be $5\sqrt{3}$ A, At some instant $I_1$ is $10\sqrt{2}$ A then at this instant I will be $10\sqrt{2}$ A

Answer 2

The circuit consists of two parallel branches connected to an AC voltage source $v(t) = 200\sqrt{2} \sin(100t) V$. The RMS voltage is $V_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 V$ and the angular frequency is $\omega = 100$ rad/s.

Branch 1: $R_A = 10 \Omega$, $L = \frac{\sqrt{3}}{10} H$.

Inductive reactance $X_L = \omega L = 100 \times \frac{\sqrt{3}}{10} = 10\sqrt{3} \Omega$.

Impedance $Z_1 = \sqrt{R_A^2 + X_L^2} = \sqrt{10^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = 20 \Omega$.

RMS current $I_{1,rms} = \frac{V_{rms}}{Z_1} = \frac{200}{20} = 10 A$.

Phase angle $\phi_1 = \arctan(\frac{X_L}{R_A}) = \arctan(\frac{10\sqrt{3}}{10}) = \arctan(\sqrt{3}) = 60^\circ$. Since it is an R-L circuit, the current lags the voltage.

Instantaneous current $i_1(t) = I_{1,0} \sin(\omega t - \phi_1) = 10\sqrt{2} \sin(100t - 60^\circ) A$.

Branch 2: $R_B = 20 \Omega$, $C = \frac{\sqrt{3}}{2} mH = \frac{\sqrt{3}}{2} \times 10^{-3} F$.

Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times \frac{\sqrt{3}}{2} \times 10^{-3}} = \frac{2 \times 10^3}{100\sqrt{3}} = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \Omega$.

Impedance $Z_2 = \sqrt{R_B^2 + X_C^2} = \sqrt{20^2 + (\frac{20\sqrt{3}}{3})^2} = \sqrt{400 + \frac{400 \times 3}{9}} = \sqrt{400 + \frac{400}{3}} = \sqrt{\frac{1600}{3}} = \frac{40}{\sqrt{3}} = \frac{40\sqrt{3}}{3} \Omega$.

RMS current $I_{2,rms} = \frac{V_{rms}}{Z_2} = \frac{200}{\frac{40\sqrt{3}}{3}} = \frac{200 \times 3}{40\sqrt{3}} = \frac{5 \times 3}{\sqrt{3}} = 5\sqrt{3} A$.

Phase angle $\phi_2 = \arctan(\frac{X_C}{R_B}) = \arctan(\frac{20\sqrt{3}/3}{20}) = \arctan(\frac{\sqrt{3}}{3}) = \arctan(\frac{1}{\sqrt{3}}) = 30^\circ$. Since it is an R-C circuit, the current leads the voltage.

Instantaneous current $i_2(t) = I_{2,0} \sin(\omega t + \phi_2) = 5\sqrt{3} \times \sqrt{2} \sin(100t + 30^\circ) = 5\sqrt{6} \sin(100t + 30^\circ) A$.

Statement 1: Phase difference between $I_1$ and $I_2$ is $90^\circ$.

The phase of $I_1$ is $-60^\circ$ and the phase of $I_2$ is $+30^\circ$ relative to the voltage.

The phase difference is $|30^\circ - (-60^\circ)| = 90^\circ$. Statement 1 is correct.

Statement 2: At some instant current in $L-R_A$ is 10 A. At the same instant current in $C-R_B$ branch will be $5\sqrt{3}$ A.

We need to check if there exists an instant $t$ such that $i_1(t) = 10 A$ and $i_2(t) = 5\sqrt{3} A$.

$10 = 10\sqrt{2} \sin(100t - 60^\circ) \implies \sin(100t - 60^\circ) = \frac{1}{\sqrt{2}}$.

$100t - 60^\circ = 45^\circ$ or $135^\circ$ (within one period).

If $100t - 60^\circ = 45^\circ$, then $100t = 105^\circ$.

$i_2(t) = 5\sqrt{6} \sin(100t + 30^\circ) = 5\sqrt{6} \sin(105^\circ + 30^\circ) = 5\sqrt{6} \sin(135^\circ) = 5\sqrt{6} \times \frac{1}{\sqrt{2}} = 5\sqrt{3} A$.

Since we found an instant where this is true, statement 2 is correct.

Statement 3: At some instant $I_1$ is $10\sqrt{2}$ A then at this instant I will be $10\sqrt{2}$ A.

$I_1$ refers to the instantaneous current $i_1(t)$. The maximum value of $i_1(t)$ is $10\sqrt{2} A$. This occurs when $\sin(100t - 60^\circ) = 1$.

So, $100t - 60^\circ = 90^\circ + 360^\circ n$, which means $100t = 150^\circ + 360^\circ n$.

At this instant, $100t + 30^\circ = 150^\circ + 30^\circ + 360^\circ n = 180^\circ + 360^\circ n$.

$i_2(t) = 5\sqrt{6} \sin(100t + 30^\circ) = 5\sqrt{6} \sin(180^\circ + 360^\circ n) = 5\sqrt{6} \times 0 = 0$.

The instantaneous main current $i(t) = i_1(t) + i_2(t)$.

At the instant when $i_1(t) = 10\sqrt{2} A$, $i_2(t) = 0$.

So, $i(t) = 10\sqrt{2} + 0 = 10\sqrt{2} A$. Statement 3 is correct.

Statement 4: Power dissipated in the circuit is 2121.3 W.

The average power dissipated in the circuit is the sum of the average power dissipated in each resistor.

Power dissipated in $R_A$: $P_1 = I_{1,rms}^2 R_A = (10)^2 \times 10 = 100 \times 10 = 1000 W$.

Power dissipated in $R_B$: $P_2 = I_{2,rms}^2 R_B = (5\sqrt{3})^2 \times 20 = (25 \times 3) \times 20 = 75 \times 20 = 1500 W$.

Total power dissipated $P = P_1 + P_2 = 1000 + 1500 = 2500 W$.

Statement 4 is incorrect.

All statements 1, 2, and 3 are correct. Since this is a multiple choice question and the provided options are checkboxes, it is implied that there can be multiple correct options.