Three resistors are connected in parallel across a DC voltage source. The current flowing through R1 and R3 are 2,6A and 3A respectively. The power drawn the second resistor is 250W and energy dissipated by the R3 in 20sec is 1800J. Calculate

Question

1) resistance of each resistor
2) supply voltage
3) current through R2
4) total resistance of the circuit

Answer 1

To analyze the circuit with three resistors (\( R_1 \), \( R_2 \), and \( R_3 \)), we can use the information given about current, power, energy, and the properties of resistors in parallel to find the unknown quantities.

1) Calculate the Resistance of Each Resistor

For \( R_3 \):

• Given that the current through \( R_3 \) is \( I_3 = 3 , \text{A} \).
• The energy dissipated by \( R_3 \) in \( 20 , \text{s} \) is \( E = 1800 , \text{J} \).
• The power (P) for \( R_3 \) can be calculated using:

\[ E = P \cdot t \implies P = \frac{E}{t} = \frac{1800 , \text{J}}{20 , \text{s}} = 90 , \text{W} \]

• Power can also be expressed as:

\[ P = I^2 R \implies R_3 = \frac{P}{I_3^2} = \frac{90 , \text{W}}{(3 , \text{A})^2} = \frac{90}{9} = 10 , \Omega \]

For \( R_2 \):

• We know the power drawn by \( R_2 \) is \( P_2 = 250 , \text{W} \).
• The relationship between voltage, power, and current gives us:

\[ P = I R \implies R_2 = \frac{P_2}{I_2} \]

• Since the voltage across all resistors in parallel is the same, we’ll find the voltage later to determine \( R_2 \).

For \( R_1 \):

• Given the current through \( R_1 \) is \( I_1 = 2.6 , \text{A} \):
• The voltage across \( R_1 \) can be found as follows. Since it is in parallel, \( V = I_1 R_1 \) gives:

\[ R_1 = \frac{V}{I_1} \]

We will come back to this after we determine the voltage.

2) Calculate the Supply Voltage

Using the power formula, we can find the voltage across \( R_2 \):

\[ P = V I \implies V = \frac{P_2}{I_2} \]

We already have \( I_3 = 3 , A \) and \( I_1 = 2.6 , A \). The total current \( I_{total} \) through the circuit is:

\[ I_{total} = I_1 + I_2 + I_3 \implies I_2 = I_{total} - (I_1 + I_3) = I_{total} - (2.6 + 3) \]

Finding Total Voltage

The total current can also be calculated in terms of total voltage \( V \):

Using \( P = V^2/R\), the voltages across resistors are equal, thus from \( P_2\):

\[ V = \sqrt{P_2 R_2} \]

Substituting \( R_2 = \frac{P_2}{V^2} \implies V = I_2 R_2 \).

But we can directly find \( V \) across \( R_3 \):

\[ V = I_3 R_3 = 3 \cdot 10 = 30 , \text{V} \]

3) Calculate the Current Through \( R_2 \)

Now we can find \( I_2 \) from \( P_2 \):

\[ P_2 = V I_2 \implies I_2 = \frac{P_2}{V} = \frac{250}{30} \approx 8.33 , \text{A} \]

4) Calculate Total Resistance of the Circuit

The total resistance \( R_{total} \) for parallel resistors is given by:

\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

First calculate \( R_1 \):

\[ R_1 = \frac{30}{2.6} \approx 11.54 , \Omega \]

We need \( R_2 \):

\[ R_2 = \frac{V^2}{P_2} = \frac{30^2}{250} = \frac{900}{250} \approx 3.6 , \Omega \]

Now substituting in the total resistance:

\[ \frac{1}{R_{total}} = \frac{1}{11.54} + \frac{1}{3.6} + \frac{1}{10} \]

Calculating the individual values:

\[ \frac{1}{R_{total}} \approx 0.0867 + 0.2778 + 0.1 = 0.4645 \]

Thus:

\[ R_{total} \approx \frac{1}{0.4645} \approx 2.15 , \Omega \]

Summary:

1. \( R_1 \approx 11.54 , \Omega \)
2. \( R_2 \approx 3.6 , \Omega \)
3. \( R_3 = 10 , \Omega \)
4. Supply voltage \( V = 30 , V \)
5. Current through \( R_2 \approx 8.33 , A \)
6. Total resistance \( R_{total} \approx 2.15 , \Omega \)