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Expert-verified Found in: Page 772 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 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: $${I_1}\left( {{r_1} + {R_1} + {R_4}} \right) + {I_2}({R_3} + {r_3} + {r_4}) - {E_1} - {E_3} + {E_4} = 0$$ ### Want to see more solutions like these? 