There is often a need to produce more than one voltage using a voltage divider.

Question

There is often a need to produce more than one voltage using a voltage divider. For example, the memory components of many personal computers require voltages of - 12 V. 6 V. and +12 V. all with respect to a common reference terminal. Select the values of R1. R2. and R3 in the circuit in Fig. P3.18 to meet the following design requirements: The total power supplied to the divider circuit by the 24 V source is 36 W when the divider is unloaded. The three voltages, all measured with respect to the common reference terminal, are 12 V. v2 = 6 V. and v3 = -12 V.

Explanation / Answer

voltage source V = 24 Volts

we have total power dissipated = 36 watts
let current in circuit be I
we have
Power = V x I
36 = 24 x I
Current I = 1.5 Amperes

Now we have, for Voltage V1:
V1 - IR1-IR2 = 0 (from kirchoff voltage law ,also common potential =0)
V1 = 1.5(R1 +R2)
V1 is given as 12 Volts
12 = 1.5 (R1+R2)
8 = R1 +R2 -----------------------------equation 1

Now we have, for Voltage V2:
V2 -IR2 = 0 (from kirchoff voltage law ,also common potential =0)
V2 = 1.5(R2)
V2 is given as 6 Volts
6 = 1.5 R2
R2 = 4 ohms

Substituting R2 = 4 ohms in equation1 , we get,
R1 = 4 ohms

Now we have, for Voltage V3:
V3 +IR3 = 0 (from kirchoff voltage law ,also common potential =0)
V3 = -IR3
V3 is given as -12 Volts
-12 = -1.5R3
R3 = 8 ohms

Therefore,
R1 = 4 ohms
R2 = 4 ohms
R3 = 8 ohms

Verification:
Current in circuit I = 24/(4+4+8) = 1.5 Amperes
Power consumed = I^2 Req = 1.5^2 x 16 = 36 Watts

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