Question

In the figure, R1 = 100Ω, R2 = R3 = 50.0 Ω, R4 = 75.0 Ω, and the ideal battery has

emf = 6.00 V. (a) what is the equivalent resistance? What is I in (b) resistance 1, (c)

resistance 2, (d) resistance 3, and (e) resistance 4?

Answer 1

Look at the circuit diagram:

We can notice that the resistors 2, 3, 4 are connected in parallel because their "upper" and "lower" terminals are connected together. That system, call it 234, is connected in series with the resistor 1.

(a) Therefore, the equivalent resistance will be

$$R=R_1+R_{234}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}=118.75\space\Omega.$$

(b) The total current is

$$I=\frac{V}{R}=0.0505\text{ A}.$$

Since R₁ and R₂₃₄ are connected in series, they all have the same current, i.e. 0.0505 A. The current in the first resistor is 0.0505 A.

(c, d) The voltage across the combination of 2,3,4 is

$$V_{234}=IR_{234}=0.947\text{ V}.$$

Since they're all in parallel, the same voltage $V_{234}$is applied to each resistor in the group. Therefore:

$$I_2=I_3=V_{234}/R_2=V_{234}/R_3=\\=0.96/50=0.0189\text{ A}.$$

(e) Like in (c) and (d):

$$I_4=V_{234}/R_4=0.947/75=0.0126\text{ A}.$$

According to Kirchhoff's current law, check whether

$$I_1=I_2+I_3+I_4:\\ 0.505=0.0189+0.0189+0.0126.$$

True. Correct.