Question

Suppose the emf of the battery in the circuit shown varies with time t so the current is given by $i(t) = 3 + 5t$ (here i is in ampere and t is in second). Take $R = 4\Omega$, $L = 6H$. The expression of the battery emf as function of time is $E(t) = mt + c$. Here m and c are constants with proper units. Find the value of $\frac{c}{m}$.

Answer 1

2.1

Answer 2

The circuit shown is a series RL circuit. According to Kirchhoff's voltage law, the sum of the voltage drops across the resistor and the inductor must be equal to the applied electromotive force (emf) of the battery.

The voltage across the resistor is given by Ohm's law: $V_R(t) = i(t)R$

The voltage across the inductor due to self-induction is given by: $V_L(t) = L\frac{di(t)}{dt}$

Therefore, the total emf of the battery $E(t)$ is the sum of these voltages: $E(t) = V_R(t) + V_L(t)$ $E(t) = i(t)R + L\frac{di(t)}{dt}$

Given values: Current $i(t) = 3 + 5t$ (A) Resistance $R = 4\Omega$ Inductance $L = 6H$

First, calculate the derivative of the current with respect to time: $\frac{di(t)}{dt} = \frac{d}{dt}(3 + 5t) = 5$ (A/s)

Now substitute the given values and the derivative into the emf equation: $E(t) = (3 + 5t)(4) + (6)(5)$ $E(t) = 12 + 20t + 30$ $E(t) = 20t + 42$ (V)

The problem states that the expression for the battery emf as a function of time is $E(t) = mt + c$. Comparing our derived expression $E(t) = 20t + 42$ with $E(t) = mt + c$: We find the constants: $m = 20$ (V/s) $c = 42$ (V)

Finally, we need to find the value of $\frac{c}{m}$: $\frac{c}{m} = \frac{42}{20}$ $\frac{c}{m} = \frac{21}{10}$ $\frac{c}{m} = 2.1$

The units of $\frac{c}{m}$ would be $\frac{V}{V/s} = s$. So the value is 2.1 seconds.