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A generator of emf 100V and with an internal resistance of 0,4ohms is connected in parallel to a battery that has an emf of 75V and internal resistance of 0,3ohms. This combination is used to supply a load that has a resistance of 35ohm. Use kirchoffs law to find:

1) the value and direction of the current through the generator
2) the value and direction of the current through the battery
3) voltage drop across the load

1 answer

To solve this problem, we will use Kirchhoff's laws. Here’s how we can analyze the circuit step by step:

1. Circuit Information

  • Generator:

    • EMF \( E_g = 100 , V \)
    • Internal Resistance \( r_g = 0.4 , \Omega \)

  • Battery:

    • EMF \( E_b = 75 , V \)
    • Internal Resistance \( r_b = 0.3 , \Omega \)

  • Load Resistance:

    • \( R_l = 35 , \Omega \)

2. Establish the Circuit Connections

The generator and battery are connected in parallel, so we need to consider the voltage across the load.

3. Analyze the Parallel Circuit with Kirchhoff's Laws

According to Kirchhoff’s voltage law, the sum of the voltage drops in a closed loop equals the sum of the EMFs (voltage sources).

Let:

  • \( I_g \) = current through the generator.
  • \( I_b \) = current through the battery.
  • \( I_l \) = total current through the load.

Since the loads are in parallel, the voltage across the load (which also equals the voltage across both the generator and battery terminals) can be expressed as:

\[ V_l = R_l \cdot I_l \]

4. Setup Current Relationships

By applying Kirchhoff's loop rule, we can write:

  1. For the generator: \[ E_g - I_g \cdot r_g - V_l = 0 \quad \Rightarrow \quad V_l = E_g - I_g \cdot r_g \]

  2. For the battery: \[ V_l - I_b \cdot r_b - E_b = 0 \quad \Rightarrow \quad V_l = I_b \cdot r_b + E_b \]

5. Relate Currents through the Load

At the load: \[ I_l = I_g + I_b \]

6. Equating Voltages Across the Load

Substituting \( V_l \) from both cases: \[ E_g - I_g \cdot r_g = I_b \cdot r_b + E_b \]

Substituting numerical values: \[ 100 - I_g \cdot 0.4 = I_b \cdot 0.3 + 75 \]

This simplifies to: \[ 25 = I_b \cdot 0.3 + I_g \cdot 0.4 \quad \text{(1)} \]

7. Write Current in Terms of Load Voltage

From \( V_l = R_l \cdot I_l \):

  1. From the generator's perspective:

    • \( V_l = 100 - I_g \times 0.4 \)

  2. From the battery's perspective:

    • \( V_l = I_b \times 0.3 + 75 \)

Both give us: \[ V_l = 100 - I_g \cdot 0.4 \quad (2) \] \[ V_l = I_b \cdot 0.3 + 75 \quad (3) \]

8. Solve the System of Equations

Using \( V_l \) from (2) in (3): \[ 100 - I_g \cdot 0.4 = I_b \cdot 0.3 + 75 \] This simplifies to: \[ 25 = I_b \cdot 0.3 + I_g \cdot 0.4 \quad (4) \]

Now we have two equations:

  1. \( I_l = I_g + I_b \)
  2. From (4), we rewrite as \( I_b = \frac{25 - 0.4 I_g}{0.3} \)

Substituting into the current relationship: \[ I_l = I_g + \frac{25 - 0.4 I_g}{0.3} \]

Multiply through by 0.3 to eliminate the fraction: \[ 0.3 I_l = 0.3 I_g + 25 - 0.4 I_g \] \[ 0.3 I_l = 25 - 0.1 I_g \quad \Rightarrow \quad I_g + 0.3 I_g = 25 \] Simplifying gives: \[ I_g = 20 \text{ A (flowing out of the generator)} \]

9. Calculating Current Through the Battery

Substituting \( I_g \) back into equation (1): \[ I_b \cdot 0.3 = 25 - 0.4(20) = 25 - 8 = 17 \] \[ I_b = \frac{17}{0.3} \approx 56.67 \text{ A (flowing into the battery)} \]

10. Find Voltage Drop Across Load

Voltage drop across the load can be found: \[ V_l = 100 - I_g \cdot 0.4 = 100 - 8 = 92 , V \]

Final answers:

  1. The current through the generator \( I_g = 20 , A \) (out of generator).
  2. The current through the battery \( I_b \approx 56.67 , A \) (into battery).
  3. The voltage drop across the load \( V_l = 92 , V \).