# B.5 Current and circuits SL Paper 1 – IBDP Physics 2025 – Exam Style Questions

IBDP Physics SL 2025 -B.5 Current and circuits SL Paper 1 Exam Style Questions

## Topic: B.5 Current and circuits SL Paper 1

Cells, Circuits, Resistivity, Ohm’s Law, Series & Parallel Circuits, Power In Circuits, Internal Resistance

## Question -B.5 Current and circuits SL Paper 1

Four identical lamps are connected in a circuit. The current through lamp L is I.

The lamps are rearranged using the same cell.

What is the current through $$L$$ ?

A. $$\frac{I}{4}$$

B. $$I$$

C. $$I$$

D. $$2 I$$

Ans:C

Current in both situation will be same as it is similar arrangement

#### Question

$$\mathrm{P}$$ and $$\mathrm{Q}$$ are two conductors of the same material connected in series. $$\mathrm{Q}$$ has a diameter twice that of $$P$$.
What is $$\frac{\text { drift speed of electrons in } P}{\text { drift speed of electrons in } Q}$$ ?

A. 4

B. 2

C. $$\frac{1}{2}$$

D. $$\frac{1}{4}$$

Ans:A

The drift speed of electrons in a conductor depends on various factors, including the current passing through the conductor and its properties. In this scenario, since conductors P and Q are connected in series, they carry the same current. The drift speed is primarily determined by the current, cross-sectional area, charge of the electron, and the number density of electrons in the material.

Let’s consider conductor P with diameter $$d$$ and conductor Q with diameter $$2d$$. The cross-sectional area ($$A$$) of the conductor is directly proportional to the square of its diameter. So, for conductor Q, the cross-sectional area is four times that of conductor P.

Given that the current is the same through both conductors and the drift speed is inversely proportional to the cross-sectional area, we can compare the drift speeds of electrons in P and Q as follows:

$\frac{\text{Drift speed in P}}{\text{Drift speed in Q}} = \frac{A_Q}{A_P} = \frac{4A_P}{A_P} = 4$

So, the drift speed of electrons in conductor P is four times that of conductor Q.

#### Question

Two resistors of equal resistance R are connected with two cells of emf e and 2e. Both cells have negligible internal resistance.

What is the current in the resistor labelled $$X$$ ?

A. $$\frac{\varepsilon}{2 R}$$

B. $$\frac{3 \varepsilon}{2 R}$$

C. $$\frac{\varepsilon}{R}$$

D. $$\frac{3 \varepsilon}{R}$$

Ans:C

Potential Difference across X

$2\varepsilon-\varepsilon=\varepsilon$

Potential diff. =current $\times R$

current $$=\frac{\varepsilon}{R}$$

#### Question

A cylindrical conductor of length $$l$$, diameter $$D$$ and resistivity $$\rho$$ has resistance $$R$$. A different cylindrical conductor of resistivity $$2\rho$$, length $$2l$$ and diameter $$2D$$ has a resistance

A.     $$2R$$

B.     $$R$$

C.     $$\frac{R}{2}$$

D.     $$\frac{R}{4}$$

### Markscheme

B

$$R=\rho \frac{L}{A}$$

$$A=\pi r^{2}$$

r=2d

$$A=\pi (2d)^{2}$$

$$A=4\pi d^{2}$$

$$R=\frac{\rho L}{4\pi d^{2}}$$

$$R=\frac{\rho l}{4\pi D^{2} }$$

$$R_{new}=\frac{(2\rho )(2l)}{4\pi (2D)^{2}}$$

$$R_{new}=R$$

#### Question

A circuit consists of a cell of electromotive force (emf) 6.0V and negligible internal resistance connected to two resistors of 4.0Ω.

The resistance of the ammeter is 1.0 Ω. What is the reading of the ammeter?

A. 2.0A

B. 3.0A

C. 4.5A

D. 6.0A

### Markscheme

A

Both  $$4\Omega$$ resistances are connected parallelly so their equivalent will be

$$\frac{4\times 4}{4+4}=2\Omega$$

Now $$2\Omega$$ and $$1\Omega$$  are in series

$$V=i\times R$$

6=i(2+1)

i=2A

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