Name: Series RL Circuit in AC Supply
Uploaded: 2019-07-19
Duration: 7 min 29 s
Description: Let's explore the current equation, power equation, and power plot for a series RL circuit with an AC supply by watching this informative video. Watch and le...

Question

In the given circuit

Calculate

(a) Current in each branch.

(b) Power generated in each resistance

(c) Total power generated in the circuit.

(d) Net current drawn from source.

(e) Net impedance of the circuit.

Accepted Answer

Given,
Resistance of first branch is, $R_{1} = 4 Ω$
Inductive reactance of first branch is, $X_{L} = 3 Ω$
Resistance of second branch is, $R_{2} = 6 Ω$
Capacitive reactance of second branch is, $X_{C} = 8 Ω$
Resistance of third branch is, $R_{3} = 10 Ω$
Inductive reactance of third branch is, $X_{L}' = 10 Ω$
Capacitive reactance is, $X_{C}' = 10 Ω$
rms voltage is, $V_{rms} = 100 V$

Total impedance of first branch is, $Z_{1} = \sqrt{R^{2} + {X_{L}}^{2}} = \sqrt{4^{2} + 3^{2}} = 5 Ω$
Total impedance of second branch is, $Z_{2} = \sqrt{R^{2} + {X_{L}}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 Ω$
Total impedance of third branch is, $Z_{3} = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{10^{2} + {(10 - 10)}^{2}} = 10 Ω$

All the branch is connected in parallel, so voltage for all the branch will be remains same, so

$(a)$ Current in first branch is, $i_{1} = \frac{V_{rms}}{Z_{1}} = \frac{100}{5} = 20 A$
Current in second branch is, $i_{2} = \frac{V_{rms}}{Z_{2}} = \frac{100}{10} = 10 A$
Current in third branch is, $i_{3} = \frac{V_{rms}}{Z_{3}} = \frac{100}{10} = 10 A$

$(b)$ Power generated in first resistance is, $P_{1} = {i_{1}}^{2} R_{1} = 20^{2} \times 4 = 1600 W$
Power generated in second resistance is, $P_{2} = {i_{2}}^{2} R_{2} = 10^{2} \times 6 = 600 W$
Power generated in third resistance is, $P_{3} = {i_{3}}^{2} R_{3} = 10^{2} \times 10 = 1000 W$

$(c)$ Total power generated in circuit is,

$P = P_{1} + P_{2} + P_{3}$
$\Rightarrow P = 1600 + 600 + 1000 = 3200 W$

$(d)$ Power factor for first branch is given by, $\cos (ϕ_{1}) = \frac{R_{1}}{Z_{1}} = \frac{4}{5}, \Rightarrow ϕ_{1} = 37 °$
Power factor for second branch is given by, $\cos (ϕ_{2}) = \frac{R_{2}}{Z_{2}} = \frac{6}{10}, \Rightarrow ϕ_{2} = 53 °$
Power factor for third branch is given by, $\cos (ϕ_{3}) = \frac{R_{3}}{Z_{3}} = \frac{10}{10}, \Rightarrow ϕ_{3} = 0 °$
Phasor diagram of circuit is given by,

From the given diagram, we can conclude,
$i_{net} = \sqrt{{(i_{1} \cos (37 °) + i_{2} \cos (53 °) + i_{3})}^{2} + {(i_{2} \sin (53 °) - i_{1} \sin (37 °))}^{2}}$
$\Rightarrow i_{net} = \sqrt{{(20 \times \frac{4}{5} + 10 \times \frac{3}{5} + 10)}^{2} + {(10 \times \frac{4}{5} - 20 \times \frac{3}{5})}^{2}}$
$\Rightarrow i_{net} = \sqrt{{(16 + 6 + 10)}^{2} + {(8 - 12)}^{2}}$
$\Rightarrow i_{net} = \sqrt{{(32)}^{2} + {(4)}^{2}} = 32.24 A$

$(e)$ Now net impedance of the circuit is, $Z = \frac{V_{rms}}{i_{net}}$
$\Rightarrow Z = \frac{100}{32.24} = 3.10 Ω$

In the given circuit

Calculate

(a) Current in each branch.

(b) Power generated in each resistance

(c) Total power generated in the circuit.

(d) Net current drawn from source.

(e) Net impedance of the circuit.

Important Questions on Alternating Current

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