The figure below employs a convention often used in circuit diagrams. The batter

Question

The figure below employs a convention often used in circuit diagrams. The battery (or other power supply) is not shown explicitly. It is understood that the point at the top, labeled "V," is connected to the terminal of a 33.8 V battery having negligible internal resistance, and that the "ground" symbol at the bottom is connected to the negative terminal of the battery. The circuit is completed through the battery, even though it is not shown on the diagram.

(a) What is the potential difference Vab, the potential of point a relative to point b, when the switch S is open?
V

(b) What is the current through switch S when it is closed?
A

(c) What is the equivalent resistance when switch S is closed?

Explanation / Answer

a) I1 = E/(R1 + R3),                 I2 = E/(R2 + R4),                   Vc = 0 volt

Va = I1*R3,                          Vb = I2*R4

Vab = Va – Vb = [E/(R1 + R3)]*R3 – [E/(R2 + R4)]*R4

                         = [33.8/(6 + 3)]*3 – [33.8/(3 + 6)]*6 = -11.27 V

b) Let I = current in whole circuit

           I1, I2, I3 and I4 are currents along R1, R2, R3 and R4 respectively.

           I5 in the loop of resistors R1 and R3.

Junction Rule:   I3 = I1 – I5,     I4 = I2 + I5

Loop rule, Loop1:   -3I2 + 3I5 + 6I1 = 0

                    Loop2:    -3I5 – 6(I2 + I5) + 3(I1 – I5) = 0

                    Loop3:    -6I1 – 3(I1 – I5) + 33.8 = 0

I1 = (3I2 – 3I5)/6 = 0.5I2 – 0.5I5

Put I1 in loop 2, we get, -13.5*I5 – 4.5*I2 = 0 ===è I2 = -3*I5

Put I2 back in I1, I1 = -2*I5

Putting I1 value in loop 3 equation, we get,   I5 = -33.8/21 = -1.61 A

c) E = Req*I,                  I = I1 + I2

        Req = E/(I1 + I2)

    I1 = -2*I5 = 3.22 A,            I2 = -3*I5 = 4.83 A

      Req = 33.8/(3.22 + 4.83) = 4.2 ohm

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