Question

Figure1 of 1The figure shows a circuit consisting of three parallel branches. 24-volt battery is the first branch. The second branch is a series combination of 5-ohm resistor and 10-ohm resistor. The third branch is a series combination of a 6-ohm resistor and 4-ohm resistor. The figure shows a circuit consisting of three parallel branches. 24-volt battery is the first branch. The second branch is a series combination of 5-ohm resistor and 10-ohm resistor. The third branch is a series combination of a 6-ohm resistor and 4-ohm resistor. Part A For the circuit shown in the figure(Figure 1) find the current through each resistor. Express your answers using two significant figures. Enter your answers numerically separated by commas. I6Ω, I5Ω, I10Ω, I4Ω = nothing A Request Answer Part B For the circuit shown in the figure find the potential difference across each resistor. Express your answers using two significant figures. Enter your answers numerically separated by commas. ΔV6Ω, ΔV5Ω, ΔV10Ω, ΔV4Ω = nothing V Request Answer Provide Feedback

Answer 1

(a)

first, we need to find equivalent capacitance

6 ohms and 4 ohms are in series, this gives 10 ohms

5 and 10 ohms are in series. this gives 15 ohms

now, this 15 ohms and 10 ohms ( from above) are in parallel

so,

R_equ = (1/15 + 1/10)^-1

R_equ = 6 ohms

so,

total current in the circuit, I = V / R_equ = 24 / 6 = 4 Amps

for resistors in parallel, we will have same voltage across them and for resistors in series, we will have same currents

_________

so,

Part A

voltage across combination of 6 ohm and 4 ohm = 24 volts

so,

I = 2.4 A

and

for combination of 5 ohm and 10 ohm,

I = 1.6 A

so,

I (6Ω) = 2.4 A

I(5Ω) = 1.6 A

I(10Ω) = 1.6 A

I(4Ω) = 2.4 A

___________________________

part B

just use V = IR

we got

ΔV6Ω = 14.4 V

ΔV5Ω = 8 V

ΔV10Ω = 16 V

ΔV4Ω = 9.6 V