Question

The charge \( q \) (in coulombs) passing through a \( 10 \Omega \) resistor as a function of time \( t \) (in seconds) is given by: \[ q = 3t^2 - 2t + 6 \] The potential difference across the ends of the resistor at time \( t = 5 \) s is:

• A. \( 120 \, V \)
• B. \( 240 \, V \)
• C. \( 140 \, V \)
• D. \( 280 \, V \)

Hint:

To find voltage across a resistor when charge is given as a function of time, first compute current using \( I = \frac{dq}{dt} \) and then use \( V = IR \).

Answer 1

The Correct Option is D

Step 1: Calculate Current Current is the time derivative of charge: \[ I = \frac{dq}{dt} = \frac{d}{dt} (3t^2 - 2t + 6) \] \[ I = 6t - 2 \] At \( t = 5 \) s: \[ I = 6(5) - 2 = 30 - 2 = 28 { A} \] Step 2: Apply Ohm’s Law \[ V = IR = 28 \times 10 = 280 { V} \] Thus, the correct answer is \( 280 \, V \).