Chapter 21: Q23CQ (page 772)

Apply the loop rule to loops \({\rm{abgefa}}\) and \({\rm{cbgedc}}\) in Figure \({\rm{21}}{\rm{.47}}\).

Short Answer

Expert verified

In loop\({\rm{abgefa}}\)the relation obtained is:\({I_1}({R_1} + {r_1} + {R_4}) + {I_3}({R_2} + {r_2}) - {E_1} - {E_2} = 0\).

In loop \({\rm{cbgedc}}\) the relation obtained is: \({I_2}({R_3} + {r_3} + {r_4}) - {I_3}({R_2} + {r_2}) + {E_2} - {E_3} + {E_4} = 0\).

Step by step solution

The Loop Rule

This rule state that if we add the potential drop across all the components in a particular loop, the result should be zero.

Evaluating the loop

Applying the rule in loop\(abgefa\), we get

\(\begin{array}{c}{E_1} - {I_1}{r_1} - {I_1}{R_1} - {I_3}{R_2} + {E_2} - {I_3}{r_2} - {I_1}{R_4} = 0\\{I_1}({R_1} + {r_1} + {R_4}) + {I_3}({R_2} + {r_2}) - {E_1} - {E_2} = 0\end{array}\)

Applying the rule in loop\(cbgedc\), we get

\(\begin{array}{c}{I_2}{R_3} - {I_3}{R_2} + {E_2} - {I_3}{r_2} + {E_4} + {I_2}{r_4} + {I_2}{r_3} - {E_3} = 0\\{I_2}({R_3} + {r_3} + {r_4}) - {I_3}({R_2} + {r_2}) + {E_2} - {E_3} + {E_4} = 0\end{array}\)

Therefore, we get:

The loop rule in\({\rm{abgefa}}\)gives the relation:\({I_1}({R_1} + {r_1} + {R_4}) + {I_3}({R_2} + {r_2}) - {E_1} - {E_2} = 0\).

The loop rule in \({\rm{cbgedc}}\) gives the relation:\({I_2}({R_3} + {r_3} + {r_4}) - {I_3}({R_2} + {r_2}) + {E_2} - {E_3} + {E_4} = 0\).