CAIE A-Level Physics AS 10.2 Kirchhoffs Laws Questions

Practise applying Kirchhoff’s laws, deriving series and parallel resistance formulae and solving multi-loop currents, potential differences and resistances.

Exam points

apply Kirchhoff’s first and second laws as conservation of charge and energy
derive and use equivalent resistance formulae for series and parallel networks
solve multi-loop circuit problems for currents, potential differences and resistances using consistent signs and junction/loop equations

Covered syllabus points

10.2.1—Kirchhoff’s first law and understand that it is a consequence of 10.2.2—Kirchhoff’s second law and understand that it is a consequence of 10.2.3—Kirchhoff’s laws, a formula for the combined resistance of two or more 10.2.4—The formula for the combined resistance of two or more resistors in series 10.2.5—Kirchhoff’s laws, a formula for the combined resistance of two or more 10.2.6—The formula for the combined resistance of two or more resistors in parallel 10.2.7—Kirchhoff’s laws to solve simple circuit problems

[Maximum number: 1]

Which description of Kirchhoff's first law is correct?

It considers the currents at a junction in a circuit and is a consequence of the conservation of charge.

It considers the currents at a junction in a circuit and is a consequence of the conservation of energy.

It considers the electromotive forces and potential differences in a circuit loop and is a consequence of the conservation of charge.

It considers the electromotive forces and potential differences in a circuit loop and is a consequence of the conservation of energy.

[Maximum number: 1]

Three resistors, $R_{1}, R_{2}$ and $R_{3}$ , are connected in parallel to a cell. The currents in the resistors are $I_{1}, I_{2}$ and $I_{3}$ . The potential differences across the resistors are $V_{1}, V_{2}$ and $V_{3}$ . The current in the cell is $I_{0}$ . The potential difference across the cell is $V_{0}$ , as shown.

Figure for Question 2 — CAIE A-Level Physics AS

Which equation can be obtained by applying Kirchhoff's second law to the circuit?

$I_{0}=I_{1}=I_{2}=I_{3}$

$I_{0}=I_{1}+I_{2}+I_{3}$

$V_{0}=V_{1}=V_{2}=V_{3}$

$V_{0}=V_{1}+V_{2}+V_{3}$

[Maximum number: 5]

(a)

State Kirchhoff's first law.

[ 1 ]

(b)

A cell with internal resistance r is connected to two resistors of resistances $R_{1}$ and $R_{2}$ as shown in Fig. 6.1.

Fig. 6.1

The potential differences (p.d.s) across $R_{1}$ and $R_{2}$ are $V_{1}$ and $V_{2}$ respectively.
The terminal p.d. across the cell is V.
The current in the circuit is I.

Use Kirchhoff's laws to show that the total resistance $R_{\mathrm{T}}$ of the external circuit is given by

R_{\mathrm{T}}=R_{1}+R_{2} .

[ 2 ]

(by Kirchhoff's second law) $V=V_{1}+V_{2}$
so $I R_{\mathrm{T}}=I R_{1}+I R_{2}($ and cancelling I gives $) R_{\mathrm{T}}=R_{1}+R_{2}$

or

$V / I=V_{1} / I+V_{2} / I$ (and substituting R gives) $R_{\mathrm{T}}=R_{1}+R_{2}$

(c)

A resistor of resistance $R_{3}$ is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2.

Fig. 6.2

State and explain the effect, if any, of this change on:

[ 2 ]

(i)

the current in the cell

[ 2 ]

[Maximum number: 2]

A cell with internal resistance r is connected to two resistors of resistances $R_{1}$ and $R_{2}$ as shown in Fig. 6.1.

Fig. 6.1

R_{\mathrm{T}}=R_{1}+R_{2} .

[Maximum number: 1]

The diagrams show two different circuits.

Figure for Question 5 — CAIE A-Level Physics AS

The cells in each circuit have the same electromotive force (e.m.f.) and negligible internal resistance. The three resistors each have the same resistance R.

In the circuit on the left, the power dissipated in the resistor is P.
What is the total power dissipated in the circuit on the right?

$\frac{P}{4}$

$\frac{P}{2}$

2 P

[Maximum number: 2]

Three resistors of resistances $R_{1}, R_{2}$ and $R_{3}$ are connected as shown in Fig. 6.1.

Fig. 6.1

The currents in the resistors are $I_{1}, I_{2}$ and $I_{3}$ . The total current in the combination of resistors is I and the potential difference across the combination is V.

Show that the total resistance R of the combination is given by the equation

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

$I=I_{1}+I_{2}+I_{3}$ $\begin{aligned} (V / R)=\left(V / R_{1}\right)+\left(V / R_{2}\right)+\left(V / R_{3}\right) \text { or } (I / V)=\left(I_{1} / V\right)+\left(I_{2} / V\right)+\left(I_{3} / V\right) \text { and (so) } 1 / R=1 / R_{1}+1 / R_{2}+1 / R_{3} \end{aligned}$

[Maximum number: 1]

A battery of electromotive force (e.m.f.) 6.0 V and internal resistance r is connected to a resistor of resistance $12 \Omega$ and a variable resistor X, as shown in Fig. 6.2.

Fig. 6.2

The combined resistance of the two resistors connected in parallel is $4.8 \Omega$ .

Calculate the resistance of X.

\text { resistance of } X=

$\begin{aligned} 1 / 4.8=1 / 12+1 / R_{\mathrm{X}} R_{\mathrm{X}}=8.0 \Omega \end{aligned}$

[Maximum number: 4]

(a)

Fig. 5.2 shows a circuit with a battery of electromotive force (e.m.f.) 12.0 V connected to a linear potentiometer AB and two identical filament lamps P and Q .

Fig. 5.2

The battery has negligible internal resistance and the lamps each have the same I-V characteristic shown in Fig. 5.1.

When the slider of the potentiometer is at its midpoint, as shown in Fig. 5.2, the current I in the battery is 1.78 A .

Determine:

[ 4 ]

(i)

the current in lamp P
current = A

[ 1 ]

$I=1.55 \mathrm{~A}$

(ii)

the resistance of the potentiometer between its ends A and B.
resistance = $\Omega$

[ 3 ]

$\begin{gathered} I=1.78-1.55 (=0.23 \mathrm{~A}) \end{gathered}$ $\begin{aligned} R =12.0 / 0.23 =52 \Omega \end{aligned}$

All question bank results loaded

Covered syllabus points

Question 1

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question 4

Question 5

Question 6

Question 7

Question 8

Question (a)

Question (i)

Question (ii)