Question
Question: A \( 100\Omega \) resistance and a capacitor of \( 100\Omega \) reactance are connected in series ac...
A 100Ω resistance and a capacitor of 100Ω reactance are connected in series across a 220V source. When the capacitor is 50percent charged, then what will be the peak value of the displacement current?
A.2.2A
B.1.41A
C.4.4A
D.112A
Solution
The value of the impedance of the circuit can be found by taking the square root of the sum of the square of resistance and the square of the reactance. The peak value of displacement current will be the ratio of the maximum voltage of the source to the impedance of the circuit. This will help you in answering this question.
Complete answer:
It has been mentioned in the question that the resistance of the resistor connected in the circuit can be written as,
R=100Ω
The reactance of the capacitor which is connected in this circuit can be mentioned as,
X=100Ω
The maximum voltage of the source connected will be given as,
V=220V
The value of the impedance of the circuit can be found by taking the square root of the sum of the square of resistance and the square of the reactance. This can be shown as an equation as,
Z=R2+X2
Substituting the values in it will give,
Z=1002+1002=1002Ω
At the condition where the capacitor which is 50percent charged, the maximum voltage can be written as,
Vmax=2110V
Therefore the peak value of displacement current will be the ratio of the maximum voltage of the source to the impedance of the circuit. That is we can write that,
I=ZVmax=10021102=2.2A
Hence the peak value of the displacement current has been obtained.
The answer has been given as option A.
Note: Impedance is defined as the measure of the total opposition provided by a circuit to current. In another way, we can say that it is the amount at which the circuit impedes the flow of charge. It is similar to resistance, but it also considers the effects of capacitance and inductance. Impedance can be expressed in ohms.