Question

what is (a ) the highest and (b) the lowest, total resistance that can be secured by the combination of four coils of resistance 4 ohm, 8 ohm, 12 ohm and 24 ohm​

Answer 1

$\text {\huge{ \underline{ \underline{Solution}}}}$

$\textbf{(a)} \: \text{The \: highest \: resistence \: can \: be \: } \\ \text{secured \: by \: connecting \: all \: the \: } \\ \text{four \: coil \: in \: series....}$

$\texttt{In \: this \: case .....}$

$R = R_1 + R_2 + R_3 + R_4 \\ \\ R = 4 + 8 + 12 + 24 \\ \\ R = 48 \omega$

$\text{Thus , \: the \: highest \: resistence \: which} \\ \text{can \: be \: secured \: is \: 48 \: ohms}.$

$\textbf{(a)} \: \text{The \: lowest \: resistence \: can \: be \: } \\ \text{secured \: by \: connecting \: all \: the \: } \\ \text{four \: coil \: in \: parallel....}$

$\texttt{In \: this \: case .....}$

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \\ \\ \frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} \\ \\ \frac{1}{R} = \frac{6 + 3 + 2 + 1}{24} \\ \\ \frac{1}{R} = \frac{12}{24} \\ \\ \frac{1}{ R} = \frac{1}{2} \\ \\ R = 2 \omega$

$\text{Thus , \: the \: lowest \: resistence \: which} \\ \text{can \: be \: secured \: is \: 2\: ohms}.$

Answer 2

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