Question

With the help of the given circuit, find out the total resistance of the circuit and the current flowing through the cell.

 

Hint:

To combine resistors in parallel, use the formula \(\frac{1}{R_{\text{parallel}}} = \sum \frac{1}{R_i}\). In series, the total resistance is simply the sum: \(R_{\text{total}} = R_1 + R_2 + ....\).

Answer 1

Given the circuit, we need to find out the total resistance and the current flowing through the cell. The circuit consists of resistors in both series and parallel combinations. 
1. Total Resistance Calculation:
- First, combine the resistors in parallel. The 500 \(\Omega\) and 750 \(\Omega\) resistors are in parallel, and the formula for the total resistance in parallel is: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{500} + \frac{1}{750} \] Simplifying this: \[ R_{\text{parallel}} = \frac{1}{\left( \frac{1}{500} + \frac{1}{750} \right)} = \frac{1}{\left( 0.002 + 0.00133 \right)} = \frac{1}{0.00333} \approx 300 \, \Omega \] 2. Total Resistance of the Circuit:
- Now, this equivalent resistance (300 \(\Omega\)) is in series with the 1000 \(\Omega\) resistor. The total resistance in series is simply the sum of the resistances: \[ R_{\text{total}} = 1000 + 300 = 1300 \, \Omega \] 3. Current Calculation:
- Using Ohm's Law, \(V = IR\), we can calculate the current flowing through the circuit. The given voltage is 4.75 V, and the total resistance is 1300 \(\Omega\): \[ I = \frac{V}{R} = \frac{4.75}{1300} \approx 0.00365 \, \text{A} = 3.65 \, \text{mA} \] Thus, the total resistance of the circuit is 1300 \(\Omega\), and the current flowing through the cell is approximately 3.65 mA.