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IBDP Physics- B.5 Current and circuits- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

In the circuit shown, the cell has negligible internal resistance.

Which equation is correct?

(A) \(I_1=4(I_2+I_3)\)
(B) \(I_1=4I_2\)
(C) \(I_3=4I_2\)
(D) \(I_3=\dfrac{4I_1}{3}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

The \(R\) resistor and the \(3R\) resistor are connected in parallel across the cell, so they have the same potential difference.

For the \(3R\) resistor,

\(V=I_2(3R)\)

For the \(R\) resistor,

\(V=I_3R\)

Since the potential difference across both resistors is the same,

\(I_3R=3RI_2\)

Therefore,

\(I_3=3I_2\)

Applying Kirchhoff’s current law at the junction,

\(I_1=I_2+I_3\)

Hence,

\(I_1=I_2+3I_2=4I_2\)

Thus, the correct equation is \( \boxed{I_1=4I_2} \).

Hence, the correct answer is \( \boxed{\mathrm{B}} \).

Question 

Three identical resistors are arranged in four different networks. The same potential difference is applied across each network. Which network dissipates the greatest electrical power?

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

The electrical power dissipated by a network connected to a fixed potential difference \(V\) is

\(P=\frac{V^2}{R_{\mathrm{eq}}}\)

Since the same potential difference is applied across every network, the network with the smallest equivalent resistance dissipates the greatest power.

Let each resistor have resistance \(R\).

For network A, all three resistors are in series:

\(R_{\mathrm{eq}}=3R\)

For network B, one resistor is in series with two parallel resistors:

\(R_{\mathrm{eq}}=R+\frac{R}{2}=\frac{3R}{2}\)

For network C, two resistors in series form \(2R\), which is in parallel with \(R\):

\(R_{\mathrm{eq}}=\frac{(2R)(R)}{2R+R}=\frac{2R}{3}\)

For network D, all three resistors are in parallel:

\(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}\)

Therefore,

\(R_{\mathrm{eq}}=\frac{R}{3}\)

Network D has the smallest equivalent resistance. Therefore, it dissipates the greatest electrical power.

Thus,

\( \boxed{\mathrm{D}} \)

Hence, the correct answer is \( \boxed{\mathrm{D}} \).

Question

Three lamps \(X\), \(Y\) and \(Z\) are connected as shown in the circuit. The emf of the cell is \(20\,\text{V}\). The internal resistance of the cell is negligible. The power dissipated by \(X\), \(Y\) and \(Z\) is \(10\,\text{W}\), \(20\,\text{W}\) and \(20\,\text{W}\) respectively.
What is the voltage across lamp \(X\) and lamp \(Y\)?
▶️ Answer/Explanation
Detailed solution

Total power supplied by the cell is the sum of the powers in the three lamps:
\( P_{\text{total}} = 10 + 20 + 20 = 50\,\text{W} \).

Since the internal resistance is negligible, the terminal voltage is \(20\,\text{V}\). The total current from the cell is
\( P_{\text{total}} = VI \Rightarrow 50 = 20I \Rightarrow I = 2.5\,\text{A} \).

Lamp \(X\) is in series with the rest of the circuit, so it carries the same current \(I=2.5\,\text{A}\). Using \(P = VI\) for lamp \(X\):
\( 10 = V_X(2.5) \Rightarrow V_X = 4\,\text{V} \).

Therefore the remaining voltage across the parallel branch containing \(Y\) and \(Z\) is
\( 20 – 4 = 16\,\text{V} \).

Lamps \(Y\) and \(Z\) are in parallel, so they each have the same potential difference:
\( V_Y = 16\,\text{V} \).

✅ Answer: (B)

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