P28_084 - i = 3 E 7 R . By the loop rule (going around the...

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84. Note that there is no voltage drop across the ammeter. Thus, the currents in the bottom resistors are thesame ,wh ichweca l l i (so the current through the battery is 2 i and the voltage drop across each of the bottom resistors is iR ). The resistor network can be reduced to an equivalence of R eq = (2 R )( R ) 2 R + R + ( R )( R ) R + R = 7 6 R which means that we can determine the current through the battery (and also through each of the bottom resistors): 2 i = E
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Unformatted text preview: i = 3 E 7 R . By the loop rule (going around the left loop, which includes the battery, resistor 2 R and one of the bottom resistors), we have E i 2 R (2 R ) iR = 0 = i 2 R = E iR 2 R . Substituting i = 3 E / 7 R , this gives i 2 R = 2 E / 7 R . The diFerence between i 2 R and i is the current through the ammeter. Thus, i ammeter = i i 2 R = 3 E 7 R 2 E 7 R = E 7 R ....
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