Question

In LCR circuit total potential is 10V and L-C-R connected in series the potential on L and C are 5v and 11 v respectively find the potential drop on R.

• A. 2
• B. 8
• C. 7
• D. 9

Hint:

In a series LCR circuit, the total voltage is the phasor sum of the voltages across the individual components. The formula \(V_{total} = \sqrt{V_R^2 + (V_L - V_C)^2}\) is essential for solving such problems, where \(V_L\) and \(V_C\) are in opposition.

Answer 1

The Correct Option is B

Step 1: Identify the given values for the LCR series circuit.
In a series LCR (Inductor-Capacitor-Resistor) circuit, the potential drops across each component are:
Potential across the inductor, \(V_L = 5\) V
Potential across the capacitor, \(V_C = 11\) V
Total potential (supply voltage), \(V_{total} = 10\) V
We need to find the potential drop across the resistor, \(V_R\). Step 2: Recall the formula for the total potential in a series LCR circuit.
In a series LCR circuit, the voltage across the resistor (\(V_R\)) is in phase with the current, while the voltage across the inductor (\(V_L\)) leads the current by 90 degrees, and the voltage across the capacitor (\(V_C\)) lags the current by 90 degrees. Therefore, \(V_L\) and \(V_C\) are 180 degrees out of phase with each other. The total potential \(V_{total}\) is the phasor sum of these individual potentials and is given by the formula: $$V_{total} = \sqrt{V_R^2 + (V_L - V_C)^2}$$ Step 3: Substitute the given values into the formula and solve for \(V_R\). We have: $$10 = \sqrt{V_R^2 + (5 - 11)^2}$$$$10 = \sqrt{V_R^2 + (-6)^2}$$$$10 = \sqrt{V_R^2 + 36}$$ To eliminate the square root, square both sides of the equation: $$10^2 = V_R^2 + 36$$ $$100 = V_R^2 + 36$$ Now, isolate \(V_R^2\): $$V_R^2 = 100 - 36$$ $$V_R^2 = 64$$ Finally, take the square root of both sides to find \(V_R\): $$V_R = \sqrt{64}$$ $$V_R = 8 \text{ V}$$ Step 4: Compare the calculated value with the given options.
The calculated potential drop across the resistor \(V_R\) is 8 V, which matches option (B).