Question

Difficulty: HardElectric Circuits and Measuring Instruments

A DC power source with an electromotive force (e.m.f.) of E=12.0 VE = 12.0\text{ V} and an internal resistance of r=1.0 Ωr = 1.0\text{ }\Omega is connected to an external load. The load consists of two parallel resistors, R1=3.0 ΩR_1 = 3.0\text{ }\Omega and R2=6.0 ΩR_2 = 6.0\text{ }\Omega, connected in series with an unknown resistor RxR_x. A real voltmeter with an internal resistance of Rv=90.0 ΩR_v = 90.0\text{ }\Omega is placed directly across the terminals of the power source and reads V=10.8 VV = 10.8\text{ V}. Calculate the resistance of RxR_x in ohms.

Answer: 8 \Omega

Answer

The resistance of RxR_x is 8.0 Ω8.0\text{ }\Omega.
Applying the relationship V=EIrV = E - Ir yields a total current of 1.2 A1.2\text{ A} through the cell. The total external resistance connected across the terminals is 9.0 Ω9.0\text{ }\Omega. Subtracting the parallel admittance of the 90.0 Ω90.0\text{ }\Omega voltmeter yields a load resistance of 10.0 Ω10.0\text{ }\Omega. Subtracting the 2.0 Ω2.0\text{ }\Omega equivalent resistance of the parallel pair (3.0 Ω3.0\text{ }\Omega and 6.0 Ω6.0\text{ }\Omega) gives 8.0 Ω8.0\text{ }\Omega for RxR_x.

Step-by-Step Solution

1
Calculate total current supplied by the cell
I=1.2 AI = 1.2\text{ A}
Terminal voltage is related to battery e.m.f. and internal resistance by V=EIrV = E - Ir.
2
Calculate equivalent resistance of the entire external circuit across terminals
Rext=9.0 ΩR_{\text{ext}} = 9.0\text{ }\Omega
By Ohm's law, Rext=VI=10.8 V1.2 A=9.0 ΩR_{\text{ext}} = \frac{V}{I} = \frac{10.8\text{ V}}{1.2\text{ A}} = 9.0\text{ }\Omega.
3
Determine the resistance of the main circuit load RLR_L
RL=10.0 ΩR_L = 10.0\text{ }\Omega
The voltmeter is in parallel with RLR_L, giving 1Rext=1Rv+1RL\frac{1}{R_{\text{ext}}} = \frac{1}{R_v} + \frac{1}{R_L}.
4
Calculate the equivalent resistance RpR_p of the parallel combination of R1R_1 and R2R_2
Rp=2.0 ΩR_p = 2.0\text{ }\Omega
Rp=R1R2R1+R2=3.0×6.03.0+6.0=2.0 ΩR_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0\text{ }\Omega.
5
Determine the value of RxR_x
Rx=8.0 ΩR_x = 8.0\text{ }\Omega
The main circuit load consists of RpR_p in series with RxR_x, so RL=Rp+RxR_L = R_p + R_x.

Key Concept

Terminal potential difference, loading effect of measuring instruments, and resistor network analysis
Estimated Time:2m 30s
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