A voltmeter measures 7.5 volts for v_o. What is the resistance of the voltmeter?

NOTE 2: I got somewhat different answers than the book. I worked the problem two different ways and got the same values, so I suspect the book is incorrect.

Although we could determine the Thevenin Equivalent by analyzing the circuit as is, I will do a small amount of circuit reduction first to reduce the number of simultaneous equations to 2. I will use mesh analysis. (Note that nodal analysis would yield 2 equations from the circuit as is, but they would be harder to derive (in my opinion) and harder to simplify.)

First, the circuit reduction. I will reduce the 16 V source and the 4 KW and 6 KW resistors to a single voltage source in series with a resistor.

NOTE: I could have reduced the circuit farther without losing i_b, but for students just learning circuit analysis, it is probably better to refrain from combining components involving parameters needed in the analysis, so I will stop the reduction here and determine the Thevenin Equivalent using mesh analysis.

First find v_oc.

First, define parameter used for dependent source in terms of mesh currents.

i_b = i₁

Next, express the value of the current source in terms of mesh currents.

0.2 i_b = - i₂

and substituting for i_b gives

0.2 i₁ = - i₂

or

0.2 i₁ + i₂ = 0 (Equation 1)

Removing the current source leaves only Mesh 1 around which to write KVL:

2.4 K i₁ + 100 i₁ + 0.4 + 10 K (i₁ - i₂) - 9.6 = 0

12.5 K i₁ - 10 K i₂ = 9.2 (Equation 2)

Solving the two equations yields

i₁ = 634.48 mA

i₂ = - 126.9 mA

Thus

v_oc = v_o = 10 K (i₁ - i₂) = 10 K (634.48 m + 126.9 m) = 7.6138 V

Next find i_sc.

Note that since the 10 KW resistor is shorted out now, it can be removed from the circuit.

Express dependent source parameters in terms of mesh currents.

i_b = i₁

Next, express the value of the current source in terms of mesh currents.

0.2 i_b = - i₂

and substituting for i_b gives

0.2 i₁ = - i₂

or

0.2 i₁ + i₂ = 0 (Equation 1)

Removing the current source leaves only Mesh 1 around which to write KVL:

2.4 K i₁ + 100 i₁ + 0.4 - 9.6 = 0

2.5 K i₁ = 9.2 (Equation 2)

Solving the two equations yields

i₁ = 3.68 mA

i₂ = - 736 mA

so

i_sc = i₁ - i₂ = 3.68 mA + 736 mA = 4.416 mA

Thus

R_Th = v_oc / i_sc = 7.6138 / 4.416 m = 1.7241 KW

and the Thevenin equivalent circuit is

Voltage division yields

7.5 = 7.6138 (R_vm) / (R_vm + 1.7241K)

7.5 (R_vm + 1.7241K) = 7.6138 R_vm

7.5 R_vm + 12.931K = 7.6138 R_vm

0.1138 R_vm = 12.931K

R_vm = 113.63 KW

What is the percentage of error in the voltage measurement?

Ideally, the voltage should equal 7.6138 V (the Thevenin Equivalent voltage). The difference in the ideal value and measured value is thus

VD = 7.5 - 7.6138 = - 0.1138 V

Thus the percentage error is

E_v = -0.1138 / 7.6138 X 100% = 1.4924%