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Problem 1. The input to this circuit is the voltage of the independent voltage source. The output is the voltage measured by the meter. Show that the output is proportional to the input. Determine the value of the constant of proportionality. Problem 2. The input to this circuit is the voltage of the independent voltage source. The output is the current measured by the meter. Show that the output is proportional to the input. Determine the value of the constant of proportionality. Problem 3 Determine the value of the voltage measured by the meter. 1 Problem 4 Determine the value of the current measured by the meter. Problem 5 Determine the value of the resistance R required to cause the value of the voltage measured by the voltmeter to be 4 V. Problem 6 The input to this circuit is the voltage of the independent voltage source, v s . The output is the current measured by the meter, i m . (a) Suppose v s = 15 V . Determine the value of the resistance R that causes the value of the current measured by the ammeter to be i m = 5 A. (b) Suppose v s = 15 V and R = 24 Ω. Determine the value of the current measured by the ammeter. (c) Suppose R = 24 Ω. Determine the value of the input voltage, v s , that causes the value of the current measured by the ammeter to be i m = 3 A. 2 Solution 1 va = 20 & 20 1 vs = vs 20 + ( 20 & 20 ) 3 3 1 12 vo = (10v a ) = × 10 × v s = 2v s 5 3 12 + 8 So vo is proportional to vs and the constant of V proportionality is 2 . V Solution 2 ib = vs 2 + ( 40 & 10 ) = vs 10 40 4 vs 4 ia = ib = = vs 40 + 10 5 10 50 100 4 8 40 io = − ( 50i a ) = − vs = − vs 3 50 3 20 + 40 The output is proportional to the input and the 8 A constant of proportionality is − . 3 V Solution 3 Replace the voltmeter with the equivalent open circuit and label the voltage measured by the voltmeter as v m . Then ib = 24 24 = =2A 4 + ( 40 & 10 ) 4 + 8 40 4 ia = i b = 2 = 1.6 A 40 + 10 5 v m = 8 i a = 8 (1.6 ) = 12.8 V 3 Solution 4 Replace the ammeter with the equivalent short circuit and label the current measured by the ammeter as i m . Then v b = 3 || ( 2 + 4 ) 3 = ( 2 ) 3 = 6 V 4 2 va = vb = 6 = 4 V 4+2 3 i m = 8 v a = 8 ( 4 ) = 32 A Solution 5 Replace the voltmeter with the equivalent open circuit and label the voltage measured by the voltmeter as v m . Then use current division in the top part of the circuit to get 40 ia = − ( 3) = −2.4 A 40 + 10 Next, use voltage division in the bottom part of the circuit to get R −5 R vm = − (5 ia ) = ia 18 + R 18 + R Combining these equations gives: 12 R −5 R vm = ( −2.4 ) = 18 + R 18 + R When vm = 4 V, 4= 12 R 18 + R ⇒ R= 4 ×18 =9 Ω 12 − 4 4 Solution 6 Replace the ammeter with the equivalent short circuit and label the current measured by the ammeter as i m . Then use voltage division in the top part of the circuit to get 2 12 va = − (vs ) = − vs 5 12 + 18 Next, use current division in the bottom part of the circuit to get 80 16 im = − (5 v a ) = − va 16 + R 16 + R Combining these equations gives: 80 2 32 im = − − vs = vs 16 + R 5 16 + R a. When vs = 15 V and im = 5 A 400 32 5= = 80 Ω 15 ⇒ 80 + 5 R = 480 ⇒ R = 5 16 + R b. When vs = 15 V and R = 24 Ω 32 im = 15 = 12 A 16 + 24 c. When im = 3 A and R = 24 Ω 4 32 3= vs = vs 5 16 + 24 ⇒ vs = 15 = 3.75 V 4 5