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: \[ I = 6(5) - 2 = 30 - 2 = 28 \text{ A} \] Step 2: Apply Ohm’s Law \[ V = IR = 28 \times 10 = 280 \text{ V} \] Thus, the correct answer is \( 280 \, V \).

Answer 2

Given:
- Charge passing through resistor, \( q = 3t^2 - 2t + 6 \) (in coulombs)
- Resistance, \( R = 10 \, \Omega \)
- Time at which potential difference is required, \( t = 5 \, \text{s} \)

Step 1: Find the current \( i \) through the resistor.
Current is the rate of change of charge with respect to time:
\[ i = \frac{dq}{dt} \] Differentiate \( q \) with respect to \( t \):
\[ i = \frac{d}{dt} (3t^2 - 2t + 6) = 6t - 2 \]

Step 2: Calculate current at \( t = 5 \) s:
\[ i = 6(5) - 2 = 30 - 2 = 28 \, \text{A} \]

Step 3: Use Ohm’s law to find potential difference \( V \):
\[ V = iR = 28 \times 10 = 280 \, V \]

Therefore, the potential difference across the resistor at \( t = 5 \) s is:
\[ \boxed{280 \, V} \]