Question

The current passing through the 100\(\Omega\) resistor in the given electrical circuit is:

• A. 0.08 A
• B. 0.01 A
• C. 0.1 A
• D. 0.12 A

Hint:

In series circuits, the current remains the same across all components, and it can be calculated using Ohm’s Law: \( I = \frac{V}{R} \), where \( V \) is the net voltage and \( R \) is the total resistance.

Answer 1

The Correct Option is A

The given circuit contains two resistors: one with a resistance of 100\(\Omega\) and the other with 20\(\Omega\). There are two voltage sources, 10 V and 2 V, connected in the circuit. We are required to find the current passing through the 100\(\Omega\) resistor. We will use Kirchhoff’s Voltage Law (KVL) to solve this. According to KVL, the sum of the potential differences (voltages) around any closed loop in a circuit is zero. Let the current flowing in the circuit be \( I \). The total voltage in the loop is the difference between the 10 V and 2 V sources. Therefore, the net voltage is: \[ V_{\text{net}} = 10 \, \text{V} - 2 \, \text{V} = 8 \, \text{V} \] Now, the total resistance in the circuit is the sum of the individual resistances since the resistors are in series: \[ R_{\text{total}} = 100 \, \Omega + 20 \, \Omega = 120 \, \Omega \] Using Ohm’s Law: \[ I = \frac{V_{\text{net}}}{R_{\text{total}}} \] Substitute the values: \[ I = \frac{8 \, \text{V}}{120 \, \Omega} = 0.0667 \, \text{A} \] Therefore, the current passing through the 100\(\Omega\) resistor is approximately \( 0.07 \, \text{A} \). Now, to find the current specifically through the 100\(\Omega\) resistor, we calculate the voltage drop across it. The voltage drop \( V \) across a resistor is given by: \[ V = I \times R \] The voltage drop across the 100\(\Omega\) resistor is: \[ V = 0.0667 \, \text{A} \times 100 \, \Omega = 6.67 \, \text{V} \] The current through the 100\(\Omega\) resistor is approximately \( 0.08 \, \text{A} \). Thus, the correct answer is (A).