3.) Two resistors, each with a resistance of are connected in parallel. Two
capacitors, each with a capacitance of are connected in series. The combination of
resistors and the combination of capacitors are then connected in series with a battery
supplying an emf of 12 V . The electrical circuit is completed. The capacitors are
uncharged at the beginning.
[70 points]
(a) Illustrate the circuit.
(b) Calculate the capacitive time constant of the circuit.
© What fraction of the final (maximum) charge is on the plates of the capacitor
combination at time ?
(d) Find an expression for the current.
(e) What fraction of the initial current remains in the circuit at ?

Ask by Reid Burton. in the United States

Mar 26,2025

Question

3.) Two resistors, each with a resistance of are connected in parallel. Two
capacitors, each with a capacitance of are connected in series. The combination of
resistors and the combination of capacitors are then connected in series with a battery
supplying an emf of 12 V . The electrical circuit is completed. The capacitors are
uncharged at the beginning.
[70 points]
(a) Illustrate the circuit.
(b) Calculate the capacitive time constant of the circuit.
© What fraction of the final (maximum) charge is on the plates of the capacitor
combination at time ?
(d) Find an expression for the current.
(e) What fraction of the initial current remains in the circuit at ?

Ask by Reid Burton. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

**(a) Circuit Diagram**

The circuit can be illustrated as follows:

```
       + 12 V battery -
        |       
        |         
   ------R₁-----  
   |             |
   |             |--> R₂  (Both resistors, each 200 MΩ, in parallel)
   |             |
   ------R₂-----  
        |
        +-------+
                |   
                |         C₁   C₂  
                +---||----+----||---+
                |         |        |
                |         |        |
                +---------+--------+
                          |
                         GND
```

In the schematic:  
- $$ R_1 $$ and $$ R_2 $$ are two resistors of $$ 200\,\mathrm{M\Omega} $$ each connected in parallel. Their equivalent resistance is  
  $$
  R_{\text{eq}} = \frac{200\,\mathrm{M\Omega}}{2} = 100\,\mathrm{M\Omega}.
  $$
- $$ C_1 $$ and $$ C_2 $$ are two capacitors of $$ 10\,\mu\mathrm{F} $$ each connected in series. Their equivalent capacitance is given by  
  $$
  \frac{1}{C_{\text{eq}}} = \frac{1}{10\,\mu\mathrm{F}} + \frac{1}{10\,\mu\mathrm{F}} = \frac{2}{10\,\mu\mathrm{F}},
  $$
  so  
  $$
  C_{\text{eq}} = 5\,\mu\mathrm{F}.
  $$

Finally, the equivalent resistor and capacitor are connected in series with the 12 V battery.

---

**(b) Capacitive Time Constant**

The time constant $$ \tau $$ for an RC circuit is given by  
$$
\tau = R_{\text{eq}} \times C_{\text{eq}}.
$$

Substitute the values:  
$$
\tau = (100 \times 10^{6}\,\Omega)(5 \times 10^{-6}\,\mathrm{F}) = 500\,\mathrm{s}.
$$

---

**(c) Fraction of Final Charge at $$ t = 100\,\mathrm{s} $$**

For a charging capacitor the charge on the capacitor as a function of time is  
$$
q(t) = q_{\text{final}} \left(1 - e^{-t/\tau}\right).
$$

The fraction of the final charge at time $$ t $$ is therefore  
$$
\frac{q(t)}{q_{\text{final}}} = 1 - e^{-t/\tau}.
$$

Plug in $$ t = 100\,\mathrm{s} $$ and $$ \tau = 500\,\mathrm{s} $$:  
$$
\frac{q(100)}{q_{\text{final}}} = 1 - e^{-100/500} = 1 - e^{-0.2}.
$$

Evaluating $$ e^{-0.2} $$ (approximately 0.8187):  
$$
\frac{q(100)}{q_{\text{final}}} \approx 1 - 0.8187 = 0.1813.
$$

Thus, about 18.13% of the final charge is on the capacitor combination at $$ t = 100\,\mathrm{s} $$.

---

**(d) Expression for the Current**

The current in an RC charging circuit is given by  
$$
i(t) = \frac{\varepsilon}{R_{\text{eq}}} e^{-t/\tau},
$$
where $$ \varepsilon = 12\,\mathrm{V} $$.

Substitute $$ R_{\text{eq}} = 100 \times 10^{6}\,\Omega $$ and $$ \tau = 500\,\mathrm{s} $$:  
$$
i(t) = \frac{12}{100 \times 10^{6}} e^{-t/500} = 1.2 \times 10^{-7}\, e^{-t/500}\,\mathrm{A}.
$$

---

**(e) Fraction of the Initial Current Remaining at $$ t = 250\,\mathrm{s} $$**

The initial current at $$ t=0 $$ is  
$$
i(0) = \frac{\varepsilon}{R_{\text{eq}}} = 1.2 \times 10^{-7}\,\mathrm{A}.
$$

At time $$ t = 250\,\mathrm{s} $$, the current is  
$$
i(250) = 1.2 \times 10^{-7}\, e^{-250/500} = 1.2 \times 10^{-7}\, e^{-0.5}\,\mathrm{A}.
$$

The fraction of the initial current remaining is  
$$
\frac{i(250)}{i(0)} = e^{-0.5}.
$$

Evaluating $$ e^{-0.5} $$ (approximately 0.6065):  
$$
\frac{i(250)}{i(0)} \approx 0.6065.
$$

Thus, approximately 60.65% of the initial current remains in the circuit at $$ t = 250\,\mathrm{s} $$.
**(a) Circuit Diagram** The circuit has two resistors ($$ 200\,\mathrm{M\Omega} $$ each) connected in parallel and two capacitors ($$ 10\,\mu\mathrm{F} $$ each) connected in series. These two combinations are then connected in series with a 12 V battery. **(b) Capacitive Time Constant** The time constant $$ \tau $$ is $$ 500\,\mathrm{s} $$. **(c) Fraction of Final Charge at $$ t = 100\,\mathrm{s} $$** Approximately 18.13% of the final charge is on the capacitors at $$ t = 100\,\mathrm{s} $$. **(d) Expression for the Current** The current $$ i(t) $$ is given by $$ i(t) = 1.2 \times 10^{-7}\, e^{-t/500}\,\mathrm{A}. $$ **(e) Fraction of Initial Current Remaining at $$ t = 250\,\mathrm{s} $$** Approximately 60.65% of the initial current remains in the circuit at $$ t = 250\,\mathrm{s} $$.

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