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College Physics (Urone)
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Short Answer

Apply the loop rule to loop \({\rm{afedcba}}\) in Figure \({\rm{21}}{\rm{.47}}\)

Applying loop rule in \({\rm{afedcba}}\) gives us: \({I_1}\left( {{r_1} + {R_1} + {R_4}} \right) + {I_2}({R_3} + {r_3} + {r_4}) - {E_1} - {E_3} + {E_4} = 0\)

See the step by step solution

Step by Step Solution

Step 1: The Loop Rule

According to this rule, the sum of potential differences of all the components in a loop should be zero,

Step 2: Evaluating the loop

Applying the loop rule in loop afedcba, we get

\(\begin{array}{l}{I_1}{r_1} - {E_1} + {I_1}{R_4} + {E_4} + {I_2}{r_4} + {I_2}{r_3} - {E_3} + {I_2}{R_3} + {I_1}{R_1} = 0\\{I_1}\left( {{r_1} + {R_1} + {R_4}} \right) + {I_2}({R_3} + {r_3} + {r_4}) - {E_1} - {E_3} + {E_4} = 0\end{array}\)

Therefore, we got the relation: \({I_1}\left( {{r_1} + {R_1} + {R_4}} \right) + {I_2}({R_3} + {r_3} + {r_4}) - {E_1} - {E_3} + {E_4} = 0\)

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