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 through the circuit using Ohm's law $(I = V / R)$:$I = \frac{10 	ext{ V}}{10 \Omega} = 1 	ext{ A}$The potential difference across the parallel combination $(V_p)$ is the current multiplied by the equivalent parallel resistance:$V_p = I 	imes R_p = 1 	ext{ A} 	imes 4 \Omega = 4 	ext{ V}$Since resistors in parallel have the same potential difference across them,the potential difference across the $12 \Omega$ resistor is $4 	ext{ V}$.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

$6 \Omega$ 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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