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Question: In an AC circuit, the potential across an inductance and resistance joined in series are respectivel...

In an AC circuit, the potential across an inductance and resistance joined in series are respectively 16V16V and 20V20V . The potential difference across the circuit is.
A. 20.0V20.0V
B. 25.6V25.6V
C. 31.9V31.9V
D. 33.6V33.6V

Explanation

Solution

In order to solve this question we need to understand AC impedance and impedance attached with inductor and capacitor. Inductor is an electric device which is used to store electrical energy in the form of a magnetic field. So when an inductor is connected in an AC circuit, then due to continuous variations in potential or current across it, it develops opposition known as inductive impedance. AC circuit is a circuit in which a sinusoidal current flows in a circuit. In this question we would first find inductive impedance and resistance and later relate with the voltage across it, finally find the total voltage.

Complete step by step answer:
Let the current in the AC circuit be, II. Also let the resistance be denoted as, RR and inductive impedance is denoted as XL{X_L}. According to the question, voltage across resistance is, VR=16V{V_R} = 16{\kern 1pt} V. And Voltage across inductor is, VL=20V{V_L} = 20V

Using Ohm's Law, we can write a relation between resistance and voltage.For Resistor, relation would be, R=VRIR = \dfrac{{{V_R}}}{I}
And for inductor it would be, XL=VLI{X_L} = \dfrac{{{V_L}}}{I}
Let the total impedance in circuit be “Z”.So from AC formula, net impedance is equal to,
Z=R2+XL2Z = \sqrt {{R^2} + {X_L}^2}
Putting values of R and XL{X_L} we get,
Z=VR2I2+VL2I2Z = \sqrt {\dfrac{{{V_R}^2}}{{{I^2}}} + \dfrac{{{V_L}^2}}{{{I^2}}}}
Z=1IVR2+VL2\Rightarrow Z = \dfrac{1}{I}\sqrt {{V_R}^2 + {V_L}^2}

Let the total voltage be, VV. So from ohm’s we law we know, Z=VIZ = \dfrac{V}{I}
Or voltage is, V=ZIV = ZI
Putting value of Z we get,
V=I×1IVR2+VL2V = I \times \dfrac{1}{I}\sqrt {{V_R}^2 + {V_L}^2}
V=VR2+VL2\Rightarrow V = \sqrt {{V_R}^2 + {V_L}^2}
Putting values we get,
V=(16)2+(20)2V = \sqrt {{{(16)}^2} + {{(20)}^2}}
V=626\Rightarrow V = \sqrt {626}
V=25.61Volt\therefore V = 25.61{\kern 1pt} Volt

So the correct option is B.

Note: It should be remembered that we have used Ohm’s law. According to Ohm’s Law, resistance developed across an electrical device is directly proportional to the voltage difference across it and inversely proportional to the current flowing through it. Also the derived values are root mean square values as the average or mean value of AC is zero.