Consider the circuit shown in the figure (Figure 1). Suppose the four resistors
ID: 1532431 • Letter: C
Question
Consider the circuit shown in the figure (Figure 1). Suppose the four resistors in this circuit have the values R_1 = 11 Ohm, R_2 = 6.3 Ohm, R_3 = 7.1 Ohm, R_4 = 10 Ohm, and that the emf of the battery is epsilon = 15 V. Find the current through each resistor using the rules for series and parallel resistors. Express your answers using two significant figures separated by commas. Find the current through each resistor using Kirchhoff's rules. Express your answers using two significant figures separated by commas.Explanation / Answer
The diagram shows a circuit with a voltage source, the four resistors, and points on the circuit labeled A, B, and C. One terminal of the voltage source is connected to point B. There are two paths from point B to point A: the first is through R3, and the second is through R2 and R4 in series. (In other words, R3 is in parallel with the series combination of R2 and R4.) Point A is connected through R1 to the other terminal of the voltage source at point C.
The first thing to do is to recognize that, since R2 and R4 are in series, their resistance adds to behave as a single resistor with value 6.3 + 10 = 16.3.
Next, since the 16.3 combination is in parallel with R3, the parallel resistance is 1/(1/16.3 + 1/7.1) 4.946
The 4.946 resistance from B to A is in series with R1, so the overall resistance is 4.946 + 11 = 15.946.
Since the battery voltage is 18V, the current through R1 = 15V/15.946 .941A and the voltage across it is .941A × 11 = 10.351V.
This means the voltage from B to A must be 15V - 10.351V = 4.65V.
The current through R3 is 4.65V/7.1 = 0.655A.
The current through R2 must equal the current through R4 because they're in series.
That's 4.65V / (6.3 + 10) = .285A
We can check that the total current from B to A is 0.655 + 0.285 = 0.94A 1A, since the current must all go through R1.