Problem 55
Question
Assertion In a series \(R-L-C\) circuit the voltage across resistor, inductor and capacitor are \(8 \mathrm{~V}, 16 \mathrm{~V}\) and \(10 \mathrm{~V}\) respectively. The resultant emf the circuit is \(10 \mathrm{~V}\). Reason Resultant emf of the circuit is given by the relation \(E=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)
Step-by-Step Solution
Verified Answer
Both the assertion and the reason are correct. The reason correctly explains the assertion.
1Step 1: Understanding the Problem
We are dealing with a RLC circuit where the voltage across the resistor, inductor, and capacitor are given. We need to check if the resultant emf matches the given emf using the formula provided.
2Step 2: Applying the Formula
According to the given formula, the resultant emf in the circuit is calculated by:\[E = \sqrt{V_R^2 + (V_L - V_C)^2}\]Substitute the given values: \(V_R = 8\, V\), \(V_L = 16\, V\), \(V_C = 10\, V\). The expression becomes:\[E = \sqrt{8^2 + (16 - 10)^2}\]
3Step 3: Calculating Values Inside the Square Root
Calculate the squares and the difference firstly:- \(8^2 = 64\)- \(16 - 10 = 6\)- \(6^2 = 36\)Now, substitute these inside the square root:\[E = \sqrt{64 + 36}\]
4Step 4: Finding the Square Root
Add the values inside the root and find the square root:\[E = \sqrt{100} = 10\, V\]
5Step 5: Conclusion and Validation
As calculated, the resultant emf in the circuit is \(10\, V\). This matches the given resultant emf. Thus, the assertion and reason are both correct, and the reason correctly explains the assertion.
Key Concepts
Voltage Across ResistorVoltage Across InductorVoltage Across CapacitorResultant EMF Calculation
Voltage Across Resistor
In an RLC circuit, understanding the voltage across a resistor is crucial. The resistor is the component that impedes the current due to resistance and this generates a voltage drop across it. In our particular exercise, the voltage across the resistor is given as 8 V.
According to Ohm's Law, the voltage drop across the resistor (\( V_R \)) is given by the formula: \[ V_R = I imes R \] where \( I \) is the current through the resistor and \( R \) is the resistance.
For this scenario, we don't need to calculate \( I \) or \( R \) because the voltage value is already supplied in the problem description. This value of 8 V is essential in determining the total emf of the circuit using the provided formula. Thus, \( V_R \)'s role is critical in calculating the overall behavior of the circuit.
According to Ohm's Law, the voltage drop across the resistor (\( V_R \)) is given by the formula: \[ V_R = I imes R \] where \( I \) is the current through the resistor and \( R \) is the resistance.
For this scenario, we don't need to calculate \( I \) or \( R \) because the voltage value is already supplied in the problem description. This value of 8 V is essential in determining the total emf of the circuit using the provided formula. Thus, \( V_R \)'s role is critical in calculating the overall behavior of the circuit.
Voltage Across Inductor
The voltage across an inductor in an RLC circuit is influenced by the rate of change of current through it. An inductor resists sudden changes in current, thereby creating a voltage drop. For the exercise, this voltage is provided as 16 V.
The induced voltage across the inductor (\( V_L \)) can be represented by: \[ V_L = L \frac{dI}{dt} \] Where \( L \) is the inductance and \( \frac{dI}{dt} \) denotes the rate of current change. However, for our exercise, we use the supplied voltage value directly to understand its impact on the total circuit.
This value enters the emf calculation as part of the difference between the inductive and capacitive voltages, which helps ascertain the circuit's resultant emf effectively.
The induced voltage across the inductor (\( V_L \)) can be represented by: \[ V_L = L \frac{dI}{dt} \] Where \( L \) is the inductance and \( \frac{dI}{dt} \) denotes the rate of current change. However, for our exercise, we use the supplied voltage value directly to understand its impact on the total circuit.
This value enters the emf calculation as part of the difference between the inductive and capacitive voltages, which helps ascertain the circuit's resultant emf effectively.
Voltage Across Capacitor
The voltage across a capacitor in an RLC circuit comes from its ability to store charge temporarily. In this case, the voltage is given as 10 V.
The voltage across the capacitor (\( V_C \)) is calculated by: \[ V_C = \frac{Q}{C} \] where \( Q \) is the electric charge on the capacitor and \( C \) is the capacitance. Alternatively, considering the alternating nature of circuits, it can also be defined by \( I \) and the capacitative reactance \( X_C \).
Given that we have the value, it is integral in computing the resultant emf by participating in the calculation involving the difference \( (V_L - V_C) \). This showcases the dynamic opposition between inductive and capacitive forces in the circuit's impedance balancing.
The voltage across the capacitor (\( V_C \)) is calculated by: \[ V_C = \frac{Q}{C} \] where \( Q \) is the electric charge on the capacitor and \( C \) is the capacitance. Alternatively, considering the alternating nature of circuits, it can also be defined by \( I \) and the capacitative reactance \( X_C \).
Given that we have the value, it is integral in computing the resultant emf by participating in the calculation involving the difference \( (V_L - V_C) \). This showcases the dynamic opposition between inductive and capacitive forces in the circuit's impedance balancing.
Resultant EMF Calculation
The resultant electromotive force (emf) in an RLC circuit represents the circuit's combined effect of voltage drops across all components. To validate if this calculated emf matches the given or observed emf, we apply the formula: \[ E = \sqrt{V_R^2 + (V_L - V_C)^2} \]
Substituting the known values from the exercise: \( V_R = 8 \, V \), \( V_L = 16 \, V \), \( V_C = 10 \, V \), The calculated emf becomes: \[ E = \sqrt{8^2 + (16 - 10)^2} \] Evaluating inside the root, we perform: - \( 8^2 = 64 \)- \( 16 - 10 = 6 \)- \( 6^2 = 36 \) Add these results: \[ \sqrt{64 + 36} = \sqrt{100} = 10 \, V \]
The resultant emf calculation confirms the given voltage for the circuit. Therefore, it illustrates the successful interaction of resistive, inductive, and capacitive elements in creating a unified circuit behavior.
Substituting the known values from the exercise: \( V_R = 8 \, V \), \( V_L = 16 \, V \), \( V_C = 10 \, V \), The calculated emf becomes: \[ E = \sqrt{8^2 + (16 - 10)^2} \] Evaluating inside the root, we perform: - \( 8^2 = 64 \)- \( 16 - 10 = 6 \)- \( 6^2 = 36 \) Add these results: \[ \sqrt{64 + 36} = \sqrt{100} = 10 \, V \]
The resultant emf calculation confirms the given voltage for the circuit. Therefore, it illustrates the successful interaction of resistive, inductive, and capacitive elements in creating a unified circuit behavior.
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