Question

In the given circuit the current flowing through the resistance 20 $\Omega$ is 0.3 A, while the ammeter reads 0.8 A. What is the value of R₁?

• A. 30 $\Omega$
• B. 40 $\Omega$
• C. 50 $\Omega$
• D. 60 $\Omega$

Answer 1

Answer: (D)

60 $\Omega$

Answer 2

The circuit shows three resistors (R₁, 20 $\Omega$, and 15 $\Omega$) connected in parallel. In a parallel circuit, the voltage across each branch is the same. The total current entering the parallel combination is the sum of the currents in each branch (Kirchhoff's Current Law).

Given:

• Current through 20 $\Omega$ resistor ($I_{20}$) = 0.3 A
• Total current ($I_{total}$) = 0.8 A

Step 1: Calculate the voltage across the 20 $\Omega$ resistor. Using Ohm's Law, $V = I \times R$: Voltage across 20 $\Omega$ resistor (V) = $I_{20} \times 20 \Omega$

$V = 0.3 A \times 20 \Omega$

$V = 6 V$

Step 2: Determine the voltage across all parallel branches. Since the resistors are in parallel, the voltage across R₁, the 20 $\Omega$ resistor, and the 15 $\Omega$ resistor is the same, which is 6 V.

Step 3: Calculate the current through the 15 $\Omega$ resistor. Using Ohm's Law, $I = V / R$: Current through 15 $\Omega$ resistor ($I_{15}$) = $V / 15 \Omega$

$I_{15} = 6 V / 15 \Omega$

$I_{15} = 0.4 A$

Step 4: Calculate the current through R₁. According to Kirchhoff's Current Law, the total current is the sum of currents in the parallel branches:

$I_{total} = I_{R1} + I_{20} + I_{15}$

$0.8 A = I_{R1} + 0.3 A + 0.4 A$

$0.8 A = I_{R1} + 0.7 A$

$I_{R1} = 0.8 A - 0.7 A$

$I_{R1} = 0.1 A$

Step 5: Calculate the value of R₁. Using Ohm's Law, $R = V / I$:

$R1 = V / I_{R1}$

$R1 = 6 V / 0.1 A$

$R1 = 60 \Omega$

The value of R₁ is 60 $\Omega$.