Question

A charged capacitor of capacitance C = 1 µF and charge Q₀ = 8 µC is connected with time varying resistance R = (5 + 3 × 10⁶ t) Ω at t = 0. The current in the circuit (in A) at t = $\frac{35}{3}$ µs is x then 10x is

Answer 1

1

Answer 2

The problem describes a discharging RC circuit where the resistance is time-varying.

1. Set up the differential equation for a discharging RC circuit:
   For a discharging capacitor, the voltage across the capacitor $V_C = Q/C$ drives the current through the resistor. The voltage drop across the resistor is $IR$.
   According to Kirchhoff's voltage law:
   $V_C - IR = 0$
   $\frac{Q}{C} = IR$

   The current $I$ is the rate at which charge leaves the capacitor, so $I = -\frac{dQ}{dt}$.
   Substituting $I$ into the equation:
   $\frac{Q}{C} = -\frac{dQ}{dt} R$
   Rearranging the terms to separate variables:
   $\frac{dQ}{Q} = -\frac{dt}{RC}$

2. Substitute the given time-varying resistance $R(t)$:
   Given $R(t) = (5 + 3 \times 10^6 t) \Omega$ and $C = 1 \mu F = 1 \times 10^{-6}$ F.
   $\frac{dQ}{Q} = -\frac{dt}{(5 + 3 \times 10^6 t) (1 \times 10^{-6})}$

3. Integrate both sides:
   Integrate from initial charge $Q_0$ at $t=0$ to charge $Q$ at time $t$:
   $\int_{Q_0}^{Q} \frac{dQ}{Q} = -\frac{1}{1 \times 10^{-6}} \int_{0}^{t} \frac{dt}{5 + 3 \times 10^6 t}$
   $\ln\left(\frac{Q}{Q_0}\right) = -10^6 \int_{0}^{t} \frac{dt}{5 + 3 \times 10^6 t}$

   To solve the integral on the right side, let $u = 5 + 3 \times 10^6 t$. Then $du = 3 \times 10^6 dt$, so $dt = \frac{du}{3 \times 10^6}$.
   When $t=0$, $u=5$. When $t=t$, $u=5+3 \times 10^6 t$.
   $\int_{0}^{t} \frac{dt}{5 + 3 \times 10^6 t} = \int_{5}^{5 + 3 \times 10^6 t} \frac{1}{u} \frac{du}{3 \times 10^6}$
   $= \frac{1}{3 \times 10^6} [\ln(u)]_{5}^{5 + 3 \times 10^6 t}$
   $= \frac{1}{3 \times 10^6} \left[\ln(5 + 3 \times 10^6 t) - \ln(5)\right]$
   $= \frac{1}{3 \times 10^6} \ln\left(\frac{5 + 3 \times 10^6 t}{5}\right)$
   $= \frac{1}{3 \times 10^6} \ln\left(1 + \frac{3 \times 10^6}{5} t\right)$

   Substitute this back into the charge equation:
   $\ln\left(\frac{Q}{Q_0}\right) = -10^6 \times \frac{1}{3 \times 10^6} \ln\left(1 + \frac{3 \times 10^6}{5} t\right)$
   $\ln\left(\frac{Q}{Q_0}\right) = -\frac{1}{3} \ln\left(1 + \frac{3 \times 10^6}{5} t\right)$
   Using logarithm properties, $a \ln x = \ln x^a$:
   $\ln\left(\frac{Q}{Q_0}\right) = \ln\left(\left(1 + \frac{3 \times 10^6}{5} t\right)^{-1/3}\right)$
   Therefore, the charge on the capacitor at time $t$ is:
   $Q(t) = Q_0 \left(1 + \frac{3 \times 10^6}{5} t\right)^{-1/3}$

4. Calculate the current $I(t)$:
   The current is $I(t) = -\frac{dQ}{dt}$.
   Differentiate $Q(t)$ with respect to $t$:
   $\frac{dQ}{dt} = Q_0 \left(-\frac{1}{3}\right) \left(1 + \frac{3 \times 10^6}{5} t\right)^{-1/3 - 1} \times \left(\frac{3 \times 10^6}{5}\right)$
   $\frac{dQ}{dt} = Q_0 \left(-\frac{1}{3}\right) \left(1 + \frac{3 \times 10^6}{5} t\right)^{-4/3} \times \left(\frac{3 \times 10^6}{5}\right)$
   $\frac{dQ}{dt} = -Q_0 \frac{10^6}{5} \left(1 + \frac{3 \times 10^6}{5} t\right)^{-4/3}$

   Now, $I(t) = -\frac{dQ}{dt}$:
   $I(t) = Q_0 \frac{10^6}{5} \left(1 + \frac{3 \times 10^6}{5} t\right)^{-4/3}$

5. Substitute the given values to find $I$ at $t = \frac{35}{3} \mu s$:
   $Q_0 = 8 \mu C = 8 \times 10^{-6}$ C
   $t = \frac{35}{3} \mu s = \frac{35}{3} \times 10^{-6}$ s

   First, calculate the term inside the parenthesis:
   $1 + \frac{3 \times 10^6}{5} t = 1 + \frac{3 \times 10^6}{5} \times \frac{35}{3} \times 10^{-6}$
   $= 1 + \frac{3 \times 35 \times 10^6 \times 10^{-6}}{5 \times 3}$
   $= 1 + \frac{105}{15}$
   $= 1 + 7 = 8$

   Now substitute this value into the current equation:
   $I = (8 \times 10^{-6}) \times \frac{10^6}{5} \times (8)^{-4/3}$
   $I = \frac{8}{5} \times (2^3)^{-4/3}$
   $I = \frac{8}{5} \times 2^{3 \times (-4/3)}$
   $I = \frac{8}{5} \times 2^{-4}$
   $I = \frac{8}{5} \times \frac{1}{2^4}$
   $I = \frac{8}{5} \times \frac{1}{16}$
   $I = \frac{1}{5 \times 2} = \frac{1}{10}$ A

   So, the current $x = 0.1$ A.

6. Calculate 10x:
   $10x = 10 \times 0.1 = 1$.