Question

A battery of 10 volt carries 20,000 C of charge through a resistance of \(20\ \Omega\). The work done in 10 seconds is

• A. \(2 \times 10^5\text{ joule}\)
• B. \(2 \times 10^3\text{ joule}\)
• C. \(2 \times 10^6\text{ joule}\)
• D. \(2 \times 10^4\text{ joule}\)

Hint:

Be careful of distractor variables!
The resistance (\(20\ \Omega\)) and time (\(10\text{ s}\)) are given to distract you.
The fundamental definition of potential difference is \(V = W/Q\), which does not depend on time or resistance.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a potential difference (\(V = 10\text{ V}\)) and a total charge (\(Q = 20,000\text{ C}\)) passing through a circuit. We need to calculate the work done by the battery.

Step 2: Key Formula or Approach:
By definition, the electric potential difference (\(V\)) between two points is the work done (\(W\)) per unit charge (\(Q\)) in moving the charge between those points:
\[ V = \frac{W}{Q} \implies W = Q \cdot V \]

Step 3: Detailed Explanation:

• Let us list the given values:
- Potential difference (\(V\)) = \(10\text{ V}\)
- Charge (\(Q\)) = \(20,000\text{ C}\)
- Resistance (\(R\)) = \(20\ \Omega\) (This is extra information not needed for the calculation).
- Time (\(t\)) = \(10\text{ s}\) (This is also extra information).

• Substitute the values of \(Q\) and \(V\) into the work formula:
\[ W = Q \cdot V \] \[ W = 20,000\text{ C} \times 10\text{ V} \] \[ W = 200,000\text{ Joules} \]

• Expressing the result in scientific exponential notation:
\[ W = 2 \times 10^5\text{ J} \]

• This matches Option (A).


Step 4: Final Answer:
The work done is \(2 \times 10^5\text{ joule}\).