Answer:
108V
Explanation:
Voltage drop across the capacitor (Vc) = IXc where
I is the current in the circuit
Xc is the capacitive reactance
Given Xc = 36Ω
To get the current I, we will use the formula for calculating the total voltage across the circuit which gives;
V = IZ
Where V is the applied voltage = 240Ω
I is the current flowing in the circuit.
Z is the impedance = 80Ω
I = V/Z
I = 240/80
I = 3A
Voltage drop across the capacitor Vc = 3×36 = 108V
The current 3A was used since the elements are connected in series and same current flows in the elements of a series connected circuit.
The voltage drop across the capacitor is 108 V
From the question,
Before we can calculate the voltage drop across the capacitor, we need to know the current flowing through the series circuit.
Applying
V = IZ................. Equation 1
Where: V = Voltage across the circuit, I = current flowing through the circuit, Z = Impendence of the circuit.
Making I the subject of the equation
I = V/Z............... Equation 2
Given: V = 240 V, Z = 80 Ω
Substitute these values into equation 2
I = 240/80
I = 3 A.
Finally using,
Vc = IXc.................... Equation 3
Where Vc = Voltage drops across the capacitor, Xc = reactance of the capacitor.
Given: Xc = 36 Ω, I = 3 A
Substitute these value into equation 3
Vc = 36(3)
Vc = 108 V
Hence, the voltage drop across the capacitor is 108 V
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