Question:
The power dissipated in the transmission cables carrying current I and voltage V is inversely proportional to
The power dissipated in the transmission cables carrying current I and voltage V is inversely proportional to
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This principle explains why electricity is transmitted at very high voltages over long distances—it minimizes the current and therefore significantly reduces energy waste as heat.
Updated On: May 6, 2026
- V
- \(V^2\)
- \(\sqrt{V}\)
- \(\sqrt{I}\)
- I
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The Correct Option is B
Solution and Explanation
Concept:
Power transmission efficiency is limited by energy loss due to resistance in cables (Joule heating).
• Transmission Power: \(P = VI\), assuming a constant power transmission at the source.
• Power Loss: The power dissipated in cables of resistance \(R_c\) is \(P_L = I^2 R_c\).
Step 1: Relate current to transmission voltage.
From the power equation \(P = VI\), the current required to transmit power \(P\) is: \[ I = \frac{P}{V} \]
Step 2: Derive the loss in terms of voltage.
Substitute the expression for \(I\) into the power loss formula: \[ P_L = \left( \frac{P}{V} \right)^2 R_c = \frac{P^2 R_c}{V^2} \]
Step 3: Identify the proportionality.
For constant transmitted power and cable resistance: \[ P_L \propto \frac{1}{V^2} \] The dissipated power is inversely proportional to the square of the voltage.
• Transmission Power: \(P = VI\), assuming a constant power transmission at the source.
• Power Loss: The power dissipated in cables of resistance \(R_c\) is \(P_L = I^2 R_c\).
Step 1: Relate current to transmission voltage.
From the power equation \(P = VI\), the current required to transmit power \(P\) is: \[ I = \frac{P}{V} \]
Step 2: Derive the loss in terms of voltage.
Substitute the expression for \(I\) into the power loss formula: \[ P_L = \left( \frac{P}{V} \right)^2 R_c = \frac{P^2 R_c}{V^2} \]
Step 3: Identify the proportionality.
For constant transmitted power and cable resistance: \[ P_L \propto \frac{1}{V^2} \] The dissipated power is inversely proportional to the square of the voltage.
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