A series combination of resistor and capacitor is connected across with an applied A.C voltage source. Separate voltmeter readings across the resistor and capacitor give value of 50 V(r.m.s) and 75V(r.m.s) respectively. The effective r.m.s voltage of the source, is nearly
a) 25 V
b) 90 V
c) 63 V
d) 125 V


Pls. show the steps.

To find the effective r.m.s voltage of the source in a series combination of a resistor and capacitor connected to an A.C. voltage source, we can use the concept of phasors and the Pythagorean theorem. In this scenario, the voltages across the resistor (V_R) and the capacitor (V_C) are given as 50 V and 75 V, respectively. Let's break down the steps to calculate the total effective voltage.

Understanding the Voltage Relationships

In an A.C. circuit with a resistor and capacitor in series, the total voltage (V) from the source is not simply the sum of the voltages across the resistor and capacitor. Instead, we treat these voltages as vectors (phasors) because they are out of phase. The voltage across the resistor (V_R) is in phase with the current, while the voltage across the capacitor (V_C) lags the current by 90 degrees.

Applying the Pythagorean Theorem

Since V_R and V_C are perpendicular to each other in the phasor diagram, we can use the Pythagorean theorem to find the total voltage:

• V = √(V_R² + V_C²)

Calculating the Total Voltage

Now, let's substitute the given values into the equation:

• V_R = 50 V
• V_C = 75 V

Now, we calculate:

• V = √(50² + 75²)
• V = √(2500 + 5625)
• V = √(8125)
• V ≈ 90.7 V

Final Result

Rounding this value, we find that the effective r.m.s voltage of the source is approximately 90 V. Therefore, the correct answer is:

• b) 90 V

This method of using the Pythagorean theorem is essential in A.C. circuit analysis, especially when dealing with components that have phase differences, such as resistors and capacitors. Understanding these relationships helps in designing and analyzing circuits effectively.