Find the currents through the three resistors shown in the figure (32-E9).
Figure32-E9
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The current via the 6 Ω and 4 Ω resistors is 0.2 A .
Explanation:
- We can observe from the figure that there will be a flow of current only in the first loop (loop 1).
- This is because of the flow of current in the slightest resistive path.
- The current will pass via the current that is linked to the wire that is in parallel in loop 2 to the 4 Ω resistor.
Therefore, in loop 2, no current will be there via the 4 Ω resistor.
Applying KVL for loop 1, we obtain:
4i + 6i + - 4 + 2 = 0
⇒10i = 2
⇒i = 0.2 A
∴ The current via the 6 Ω and 4 Ω resistors, i = 0.2 A
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