An electric circuit consists of three resistors of resistances $R1 = 4 \; \Omega$, \R2 = 6 \; \Omega\), and $R3 = 12 \; \Omega$ connected in parallel across a 12V battery. (a) Calculate the equivalent resistance of the parallel combination. (b) Calculate the electric current flowing through each resistor. (c) Calculate the total current in the circuit. (d) What is the power consumed by the 12 $\Omega$ resistor?

Q: An electric circuit consists of three resistors of resistances $R1 = 4 \; \Omega$, \R2 = 6 \; \Omega\), and $R3 = 12 \; \Omega$ connected in parallel across a 12V battery. (a) Calculate the equivalent resistance of the parallel combination. (b) Calculate the electric current flowing through each resistor. (c) Calculate the total current in the circuit. (d) What is the power consumed by the 12 $\Omega$ resistor?

Solution for Parallel Combination Circuit Given Data: Voltage $(V)$ = 12 V Resistance 1 $(R1)$ = 4 $\Omega$ Resistance 2 $(R2)$ = 6 $\Omega$ Resistance 3 $(R3)$ = 12 $\Omega$ --- (a) Equivalent Resistance of the Parallel Combination\nThe formula for equivalent resistance $(Rp)$ in parallel is: \\frac{1}{Rp} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}\\nSubstitute the values: \\frac{1}{Rp} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\\nFind the common denominator (which is 12): \\frac{1}{Rp} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{3 + 2 + 1}{12} = \frac{6}{12} = \frac{1}{2}\\nTherefore, $Rp = 2 \; \Omega$. --- (b) Electric Current Flowing Through Each Resistor\nIn a parallel circuit, the potential difference across each resistor is the same and equal to the supply voltage $(V = 12\text{ V})$. Current through $R1$ $(I1)$: \I1 = \frac{V}{R1} = \frac{12}{4} = 3 \text{ A}\ Current through $R2$ $(I2)$: \I2 = \frac{V}{R2} = \frac{12}{6} = 2 \text{ A}\ Current through $R3$ $(I3)$: \I3 = \frac{V}{R3} = \frac{12}{12} = 1 \text{ A}\ --- (c) Total Current in the Circuit\nThe total current $(I{\text{total}})$ is the sum of currents through individual resistors: \I{\text{total}} = I1 + I2 + I3 = 3 + 2 + 1 = 6 \text{ A}\ (Alternatively, using Ohm's Law with equivalent resistance: $I{\text{total}} = \frac{V}{Rp} = \frac{12}{2} = 6 \text{ A}$.) --- (d) Power Consumed by the 12 $\Omega$ Resistor\nThe formula for electric power $(P)$ is: \P = \frac{V^2}{R3} \quad \text{or} \quad P = I3^2 \times R3\\nUsing $P = I3^2 \times R3$: \P = (1)^2 \times 12 = 1 \times 12 = 12 \text{ Watts (W)}\)

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Accepted Answer

Solution for Parallel Combination Circuit

Given Data:

Voltage ((V)) = 12 V
Resistance 1 ((R_1)) = 4 (\Omega)
Resistance 2 ((R_2)) = 6 (\Omega)
Resistance 3 ((R_3)) = 12 (\Omega)

(a) Equivalent Resistance of the Parallel Combination\nThe formula for equivalent resistance ((R_p)) in parallel is:

[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}]\nSubstitute the values: [\frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}]\nFind the common denominator (which is 12): [\frac{1}{R_p} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{3 + 2 + 1}{12} = \frac{6}{12} = \frac{1}{2}]\nTherefore, (R_p = 2 ; \Omega).

(b) Electric Current Flowing Through Each Resistor\nIn a parallel circuit, the potential difference across each resistor is the same and equal to the supply voltage ((V = 12\text{ V})).

Current through (R_1) ((I_1)): [I_1 = \frac{V}{R_1} = \frac{12}{4} = 3 \text{ A}]
Current through (R_2) ((I_2)): [I_2 = \frac{V}{R_2} = \frac{12}{6} = 2 \text{ A}]
Current through (R_3) ((I_3)): [I_3 = \frac{V}{R_3} = \frac{12}{12} = 1 \text{ A}]

(c) Total Current in the Circuit\nThe total current ((I_{\text{total}})) is the sum of currents through individual resistors:

[I_{\text{total}} = I_1 + I_2 + I_3 = 3 + 2 + 1 = 6 \text{ A}] (Alternatively, using Ohm's Law with equivalent resistance: (I_{\text{total}} = \frac{V}{R_p} = \frac{12}{2} = 6 \text{ A}).)

(d) Power Consumed by the 12 (\Omega) Resistor\nThe formula for electric power ((P)) is:

[P = \frac{V^2}{R_3} \quad \text{or} \quad P = I_3^2 \times R_3]\nUsing (P = I_3^2 \times R_3): [P = (1)^2 \times 12 = 1 \times 12 = 12 \text{ Watts (W)})