In the electrical circuit shown in the figure | vedclass

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(D) $1$. First,simplify the circuit. The $20 \; \Omega$ resistor and the $30 \; \Omega$ resistor are in series with each other. Their equivalent resis

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$\frac{6}{25} \; A$

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(D) $1$. First,simplify the circuit. The $20 \; \Omega$ resistor and the $30 \; \Omega$ resistor are in series with each other. Their equivalent resistance is $R_s = 20 \; \Omega + 30 \; \Omega = 50 \; \Omega$.$2$. This $50 \; \Omega$ combination is in parallel with the wire connecting the right side of the circuit. However,looking at the circuit,the $20 \; \Omega$ resistor is in series with the $30 \; \Omega$ resistor,and this branch is in parallel with the wire segment. Wait,let's re-evaluate: The $20 \; \Omega$ resistor is in series with the $30 \; \Omega$ resistor. This branch is in parallel with the wire connecting the top and bottom nodes. Actually,the circuit shows the $20 \; \Omega$ resistor is between $A$ and $B$. The $30 \; \Omega$ resistor is in series with the wire segment. Let's simplify: The total resistance $R_{eq} = 10 \; \Omega + 3 \; \Omega + (20 \; \Omega \parallel 30 \; \Omega)$.$3$. $R_p = \frac{20 	imes 30}{20 + 30} = \frac{600}{50} = 12 \; \Omega$.$4$. Total resistance $R_{eq} = 10 + 3 + 12 = 25 \; \Omega$.$5$. Total current $I = \frac{V}{R_{eq}} = \frac{10 \; V}{25 \; \Omega} = 0.4 \; A$.$6$. Using the current divider rule to find the current $i$ through the $20 \; \Omega$ resistor: $i = I 	imes \frac{30}{20 + 30} = 0.4 	imes \frac{30}{50} = 0.4 	imes 0.6 = 0.24 \; A = \frac{24}{100} \; A = \frac{6}{25} \; A$.

In the electrical circuit shown in the figure,the current $i$ through the side $AB$ containing the $20 \; \Omega$ resistor is:

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