6 Omega and 12 Omega resistors are connected | vedclass

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(A) First, calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $12 \Omega$ resistors:$R_p = \frac{6 	imes 12}{6 + 12} =

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$4$

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(A) First, calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $12 \Omega$ resistors:$R_p = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \Omega$Next, calculate the total resistance of the circuit, as the parallel combination is in series with the $6 \Omega$ resistor:$R_{eq} = R_p + 6 \Omega = 4 \Omega + 6 \Omega = 10 \Omega$Now, calculate the total current flowing from the $10 	ext{ V}$ battery:$I = \frac{V}{R_{eq}} = \frac{10 	ext{ V}}{10 \Omega} = 1 	ext{ A}$The potential difference across the parallel combination is the product of the total current and the equivalent resistance of the parallel part:$V_p = I 	imes R_p = 1 	ext{ A} 	imes 4 \Omega = 4 	ext{ V}$Since the $6 \Omega$ and $12 \Omega$ resistors are connected in parallel, the potential difference across each of them is the same and is equal to the potential difference across the parallel combination.Therefore, the potential difference across the $12 \Omega$ resistor is $4 	ext{ V}$.

$6 \Omega$ and $12 \Omega$ resistors are connected in parallel. This combination is connected in series with a $10 \text{ V}$ battery and a $6 \Omega$ resistor. What is the potential difference between the terminals of the $12 \Omega$ resistor (in $\text{ V}$)?

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