Question

A source of e.m.f. \[E = 15\,V\] and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as \[I = 1.2t + 3\]. Then total charge will flow in first five seconds will be:
A. 10 C
B. 20 C
C. 30 C
D. 40 C

Answer 1

Hint:The electric current is the rate of flow of charge per unit time through a cross-section. So, if the flow of the charge is a function of time, then to find the electric current we find the derivative of the charge function with respect to time.

Formula used:
\[i = \dfrac{{dq}}{{dt}}\]
where i is the electric current and \[\dfrac{{dq}}{{dt}}\] is the rate of flow of charge.

Complete step by step solution:
The source emf is constant and equal to 15 volts. The resistance is changing with time and hence the electric current will also change with time.The electric current is given as the function of time,
\[i = 1.2t + 3\]
Using the formula of electric current,
\[i = \dfrac{{dq}}{{dt}}\]
\[\Rightarrow dq = idt\]

On integrating the expression, we get the charge
\[\int_{{q_{t\left( 2 \right)}}}^{{q_{t\left( 2 \right)}}} {dq} = \int_{{t_1}}^{{t_2}} {idt} \\ \]
\[\Rightarrow Q\left( {{t_2}} \right) - Q\left( {{t_1}} \right) = \int_{{t_1}}^{{t_2}} {idt} \\ \]
\[\Rightarrow Q = \int_{{t_1}}^{{t_2}} {idt} \]
We need to find the charge flow in first five seconds, so the initial time is the time when the measurement started \[{t_1} = 0s\] and the final time is the time within which the total charge needs to be calculated \[{t_2} = 5s\].

On evaluating the integral, we get
\[Q = \int_0^5 {\left( {1.2t + 3} \right)dt} \\ \]
\[\Rightarrow Q = \int_0^5 {\left( {1.2t} \right)dt} + \int_0^5 {\left( 3 \right)dt} \\ \]
\[\Rightarrow Q = 1.2\left[ {\dfrac{{{t^2}}}{2}} \right]_0^5 + 3\left[ t \right]_0^5 \\ \]
On simplifying, we get
\[Q = 1.2\left[ {\dfrac{{{5^2}}}{2} - \dfrac{{{0^2}}}{2}} \right] + 3\left[ {5 - 0} \right] \\ \]
\[\Rightarrow Q = \left( {15 + 15} \right)C \\ \]
\[\therefore Q = 30C\]
Hence, a total 30C charge will flow in the circuit in the first five seconds.

Therefore, the correct option is C.

Note: The electric charge flows continuously through the resistor when a potential difference is applied across it. So, to find the total charge flown we integrate the function of the electric current with respect to time.