Question
Question: In a parallel RC circuit, there is \(100\,mA\) through the resistive branch and \(100\,mA\) through ...
In a parallel RC circuit, there is 100mA through the resistive branch and 100mA through the capacitive branch. Find the total rms current.
A. 200mA
B. 100mA
C. 282mA
D. 141mA
Solution
Hint- For a parallel RC circuit the resistance and capacitor will be connected parallel to each other. Let the current through the resistive branch be IR and current through the capacitive branch be IC .Total current I splits into IR and IC. The relation between RMS current and peak current is given as
Irms=2I0
Where Irms is RMS current and I0 is the peak current
Using this we can find the value of total rms current in the given circuit.
Step by step solution:
A parallel RC circuit is a circuit in which the resistance and capacitor are connected parallel to each other. Since they are connected in parallel voltage across both resistor and capacitor will be equal.
But the current flowing through the resistive branch and capacitive branch will be different.
Given the current through the resistive branch is
IR=100mA
Current through the capacitive branch is
IC=100mA
From the figure we can see that total current I splits into IR and IC .
hence total current in the circuit can be found out by adding current in the resistive branch and current in the capacitive branch.
Thus, we can write,
I=IR+IC
Let us substitute the value of current in the resistive branch and capacitive branch in the above equation to find the total current.
I=100mA+100mA
∴I=200mA
We are asked to find the RMS current.
The relation between RMS current and peak current is given as
Irms=2I0
Where Irms is RMS current and I0 is the peak current
On substituting the value of peak current, we get
Irms=2200mA
∴Irms=141mA
So, the answer is option D
Note: In the question RMS value of current is asked. When we add the given values of current through resistor and current through capacitor what we get is the peak current always remember to convert it into RMS current by dividing it by 2 .