To solve the problem step by step, we will follow these steps:
### Step 1: Understand the Circuit Configuration
We have a battery of 10V connected to a fixed resistor of 20Ω and a variable resistor \( R \). The variable resistance \( R \) is increasing at a rate of \( 5 \, \Omega/\text{min} \).
### Step 2: Define the Variable Resistance
Let the variable resistance \( R \) at time \( t \) (in minutes) be given by:
\[
R(t) = 5t \, \Omega
\]
where \( t \) is the time in minutes.
### Step 3: Calculate the Total Resistance
The total resistance \( R_{\text{total}} \) in the circuit is the sum of the fixed resistance and the variable resistance:
\[
R_{\text{total}} = 20 \, \Omega + R(t) = 20 \, \Omega + 5t \, \Omega = 20 + 5t \, \Omega
\]
### Step 4: Calculate the Current in the Circuit
Using Ohm's Law, the current \( I \) in the circuit can be calculated as:
\[
I = \frac{V}{R_{\text{total}}} = \frac{10 \, \text{V}}{20 + 5t}
\]
### Step 5: Relate Current to Charge
The charge \( dq \) that passes through the battery in a small time \( dt \) can be expressed as:
\[
dq = I \, dt = \frac{10}{20 + 5t} \, dt
\]
### Step 6: Integrate to Find Total Charge
To find the total charge \( Q \) that passes through the battery in 4 minutes, we need to integrate \( dq \) from \( t = 0 \) to \( t = 4 \):
\[
Q = \int_0^4 \frac{10}{20 + 5t} \, dt
\]
### Step 7: Solve the Integral
First, simplify the integral:
\[
Q = 10 \int_0^4 \frac{1}{20 + 5t} \, dt
\]
Let \( u = 20 + 5t \), then \( du = 5 \, dt \) or \( dt = \frac{du}{5} \). Changing the limits:
- When \( t = 0 \), \( u = 20 \)
- When \( t = 4 \), \( u = 20 + 20 = 40 \)
Now, substitute into the integral:
\[
Q = 10 \int_{20}^{40} \frac{1}{u} \cdot \frac{du}{5} = 2 \int_{20}^{40} \frac{1}{u} \, du
\]
\[
Q = 2 [\ln u]_{20}^{40} = 2 (\ln 40 - \ln 20) = 2 \ln \left(\frac{40}{20}\right) = 2 \ln 2
\]
### Step 8: Convert to SI Units
Since we calculated the charge in terms of minutes, we need to convert it to seconds. The total time is \( 4 \, \text{min} = 240 \, \text{s} \):
\[
Q = 2 \ln 2 \times 60 \, \text{s}
\]
### Final Calculation
Calculating \( Q \):
\[
Q = 120 \ln 2 \, \text{C}
\]
### Conclusion
The amount of charge that passes through the battery in 4 minutes is \( 120 \ln 2 \, \text{C} \).
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