(In Parallel circuit) In procedure 2, trial 1, if you measure the following valu
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Question
(In Parallel circuit) In procedure 2, trial 1, if you measure the following values: Resistor # Current (mA) Voltage(V)
1 15.6 6.01
2 10.1 6.01
3 8.43 6.01
What is the total resistance of the circuit? Enter your response to three significant figures and in units of .
In procedure 1, trial 1, if you measure the following values: Resistor # Current (mA) Voltage (V)
1 10.2 1.51
2 10.2 2.04
3 10.2 2.45
(Series circuit) What is the total resistance of the circuit? Enter your response to three significant figures and in units of .
In procedure 1, trial 1, if you measure the following values ( in Series circuit):
What is the total resistance of the circuit? Enter your response to three significant figures and in units of .
Resistor # Current (mA) Voltage (V) 1 16.4 1.51 2 16.4 2.04 3 16.4 2.45Explanation / Answer
In parallel combination ,the potential drop across each resistor is same
Net resistance is R_total
1/R_total = (1/R1)+(1/R2)+(1/R3)
R1 = V1/i1 = 6.01/(15.6*10^-3) = 385.25 ohm
R2 = V2/i2 = 6.01/(10.1*10^-3) = 595.05 ohm
R3 = V3/i3 = 6.01/(8.43*10^-3) = 712.93 0hm
1/R_total = (1/R1)+(1/R2)+(1/R3)
1/R_total = (1/385.25)+(1/595.05)+(1/712.93)
R_total = 176.1 ohm
In series combination ,the current passing through every wire is same
R_total = R1+R2+R3
R1 = V1/I1 = 1.51/(16.4*10^-3) = 92.07 ohm
R2 = V2/I2 = 2.04/(16.4*10^-3) = 124.4 ohm
R3 = V3/I3 = 2.45/(16.4*10^-3) = 149.4 ohm
R_total = R1+R2+R3 = 92.07+124.4+149.4 = 365.87 ohm
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