Question

The current in a conductor is expressed as I = 3t² + 4t³ , where I is in Ampere and t is in second. The amount of electric charge that flows through a section of the conductor during t = 1s to t = 2s is ____________ C.

Answer 1

Correct Answer: 22

The problem provides an expression for current as a function of time, \( I = 3t^2 + 4t^3 \), and asks for the total electric charge that flows through a conductor over the time interval from t = 1s to t = 2s.

Concept Used:

Electric current \(I\) is defined as the rate of flow of electric charge \(Q\). Mathematically, this relationship is expressed as:

\[ I = \frac{dQ}{dt} \]

To find the total charge \(Q\) that flows through a cross-section of the conductor during a specific time interval from \(t_1\) to \(t_2\), we need to integrate the current \(I(t)\) with respect to time over that interval:

\[ Q = \int_{t_1}^{t_2} I(t) \, dt \]

Step-by-Step Solution:

Step 1: Identify the given expression for current and the time interval.

The current is given by: \[ I(t) = 3t^2 + 4t^3 \, (\text{in Amperes}) \] The time interval is from \( t_1 = 1 \, \text{s} \) to \( t_2 = 2 \, \text{s} \).

Step 2: Set up the definite integral for the total charge \(Q\).

Using the formula for total charge, we substitute the expression for \(I(t)\) and the limits of integration:

\[ Q = \int_{1}^{2} (3t^2 + 4t^3) \, dt \]

Step 3: Evaluate the integral.

We integrate the expression term by term using the power rule for integration, \( \int t^n \, dt = \frac{t^{n+1}}{n+1} \).

\[ \int (3t^2 + 4t^3) \, dt = 3 \int t^2 \, dt + 4 \int t^3 \, dt \] \[ = 3 \left( \frac{t^3}{3} \right) + 4 \left( \frac{t^4}{4} \right) = t^3 + t^4 \]

Step 4: Apply the limits of integration to find the value of the definite integral.

\[ Q = [t^3 + t^4]_{1}^{2} \]

Substitute the upper limit (t=2) and the lower limit (t=1) and subtract:

\[ Q = (2^3 + 2^4) - (1^3 + 1^4) \]

Final Computation & Result:

Calculate the values for each term:

\[ Q = (8 + 16) - (1 + 1) \] \[ Q = 24 - 2 \] \[ Q = 22 \, \text{C} \]

The amount of electric charge that flows through a section of the conductor during t = 1s to t = 2s is 22 C.

Answer 2

The problem provides an expression for the current \( I \) as a function of time \( t \), and asks for the total amount of electric charge that passes through a section of the conductor over a specific time interval.

Concept Used:

Electric current \( I \) is defined as the rate of flow of electric charge \( Q \). Mathematically, this relationship is expressed as:

\[ I = \frac{dQ}{dt} \]

To find the total charge \( Q \) that flows through a conductor over a time interval from \( t_1 \) to \( t_2 \), we need to integrate the current \( I(t) \) with respect to time over this interval:

\[ Q = \int_{t_1}^{t_2} I(t) \,dt \]

Step-by-Step Solution:

Step 1: Identify the given information and set up the integral.

The current is given by the expression: \( I(t) = 3t^2 + 4t^3 \) Amperes.

The time interval is from \( t_1 = 1 \) s to \( t_2 = 2 \) s.

The total charge \( Q \) is the definite integral of the current function over this interval:

\[ Q = \int_{1}^{2} (3t^2 + 4t^3) \,dt \]

Step 2: Evaluate the integral.

We integrate the expression term by term using the power rule for integration, \( \int x^n \,dx = \frac{x^{n+1}}{n+1} \):

\[ \int (3t^2 + 4t^3) \,dt = 3 \int t^2 \,dt + 4 \int t^3 \,dt \] \[ = 3 \left( \frac{t^3}{3} \right) + 4 \left( \frac{t^4}{4} \right) = t^3 + t^4 \]

Step 3: Apply the limits of integration to find the definite integral.

Now, we evaluate the antiderivative at the upper and lower limits:

\[ Q = \left[ t^3 + t^4 \right]_{1}^{2} \] \[ Q = (2^3 + 2^4) - (1^3 + 1^4) \]

Final Computation & Result:

Perform the final calculation:

\[ Q = (8 + 16) - (1 + 1) \] \[ Q = 24 - 2 \] \[ Q = 22 \text{ C} \]

The amount of electric charge that flows through the conductor during the given time interval is 22 C.