1. Which of the following must be true of the currents in the circuit? i A = i B

Question

1. Which of the following must be true of the currents in the circuit?

iA = iB, only

iA = iC, only

iA = iB = iC

iA = iB + iC

iA = iB iC

None of these.

Explain your reasoning including which Kirchhoff principle you used.

2. What is the relationship between iB and iC?

iB = 1/3 iC

iB = 1/2 iC

iB = iC

iB = 2iC

iB = 3iC

None of these but something else.

There is not enough information to decide.

Explain your reasoning including which Kirchhoff principle you used.

3. Which of the following must be true about the potential drops in the circuit?

VA = VC

VB = VC

VA = VB

V0 = VB + VC

V0 = VA + VB

V0 = VA + VC

Explain your reasoning including which Kirchhoff principle you used.

In the circuit shown at the right, RA is identical to B, and their resistance is half of RC. That is, RA = RB = ½ RC. The current through resistor A is iA, iB is the current through resistor B, and iC through resistor C. The potential drop across resistor A is VA and so on. The battery provides an EMF = V0. For each question, select all the correct answers and explain your reasoning in the space below. 1C Rc

Explanation / Answer

a)

IA=IB+IC

From Kirchoff's Junction rule ,the currents entering into the juction at the top node should be equal current leaving the junction at the top of the node

b)

equivalent resistance

Req=(RC/2) +(1/(RC/2) +1/RC)-1=0.8333RC

Current through RA is

IA=V/0.8333RC =1.2V/RC

IB=(1.2V/RC)[RC/(0.5RC+RC)]=0.8V/RC

IC=(1.2/RC)[0.5RC/(0.5RC+RC)]=0.4V/RC

Therefore

IB=2IC

c)

VB=IBRB=(0.8V/RC)(RC/2) =0.4V

VC=ICRC=(0.4V/RC)(RC)=0.4V

Therefore

VB=VC

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.