In this example we will use the concept of equivalent resistance to determine th

Question

In this example we will use the concept of equivalent resistance to determine the current in a circuit. Three identical resistors with resistances of 6.0 Ohm are connected as shown in (Figure 1) to a battery with an emf of 18.0 V and zero internal resistance. (a) Find the equivalent resistance of the resistor network. (b) Find the current in each resistor. The voltage across the parallels combination of R_2 and R_3 is the difference between joints and d, which is V_23 = 6.0 V. Since the voltage across R_2 is 6.0 V, we can the current I_2 through R_2: I_2 = 6.0 V/6.0 Ohm = 1.0 A. A similar calculation gives the current I_2 through R_3 I_3 = 6.0 V/6.0 Ohm = 1.0 A. In summary, I_1 = 2.0 A and I_2 = I_2 = 1.0 A as shown in (Figure 1). The two identical resistors R_2 and R_1 form a parable continuation the 2.0 A current arriving at point C in circuit divides equal: Hat goes through R_2 and half through R_3. At point of these two 1.0 A currents recombine into a 2.0 A current. what is the current through R_1 in the network in this example if the equal but instead R_1 = 3.0 Ohm, R_2 = 4.0 Ohm, R_3 = 2.0 Ohm?

Explanation / Answer

1) R2 and R3 are in parallel.

1/R23 = 1/R2 + 1/R3 = 1/6 + 1/6

R23 = 3 ohms

R1 = 6 ohms

Since R1 and R23 are in series

Req = R1 + R23 = 6 + 3 = 9 ohms

Using potential drop method,

V - I1R1 - I1R23 = 0

18 - I1x6 - I1x3 = 0

I1 = 2 A

This is the current through R1

Voltage at R23, V23 = V - I1R1 = 18 - 2x6 = 18 - 12 = 6 V

Since R2 and R3 are in parallel, voltage across them will be same.

V2 = V3 = V23 = 6V

V2 = I2R2 = I2x6

I2 = 6V/6ohms = 1 A

Similarly, I3 = 1 A

2) Now we have R1 = 3 ohms, R2 = 4 ohms and R3 = 2 ohms

Since R2 and R3 are in parallel,

1/R23 = 1/R2 + 1/R3 = 1/4 + 1/2

R23 = 4/3 ohms

R1 and R23 are in series,

Req = R1 + R23 = 3 + 4/3 = 13/3 ohms

Using potential drop method,

V - I1R1 - I1R23 = 0

18 - I1x3 - I1x (4/3) = 0

I1 = 18/(3+4/3) = 4.15 A

This is the current through R1.

3) Voltage at R23, V23 = V - I1R1 = 18 - (4.15x3) = 5.53 V

Since R2 and R3 are in parallel,

V2 = V3 = V23

V2 = I2R2 = I2x4

5.53 = I2 x 4

I2 = 5.53/4 = 1.38 A

This is the current through R2.

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