In the figure shown below, a resistance of 150.4Ω$150.4\Omega$ is connected in series to an ammeter A of resistance 240Ω$240\Omega$. A shunt resistance of 10Ω$10\Omega$ is connected in parallel with the ammeter. The reading of the ammeter is ________ mA$\mathrm{mA}$.
A circuit schematic containing a 20V battery connected in series with a 150.4 ohm resistor and a parallel combination of a 240 ohm ammeter and a 10 ohm shunt resistor.
Numerical Answer Type:
Enter a numerical valueAnswer: 5 to 5+4 marks
Solution & Explanation
Related Formula
Parallel combination of ammeter (RG = 240Ω$R_G = 240\Omega$) and shunt (S = 10Ω$S = 10\Omega$):
Therefore, the reading of the ammeter is 5~mA$5\mathrm{~mA}$.
A circuit schematic containing a 20V battery connected in series with a 150.4 ohm resistor and a parallel combination of a 240 ohm ammeter and a 10 ohm shunt resistor.A circuit schematic containing a 20V battery connected in series with a 150.4 ohm resistor and a parallel combination of a 240 ohm ammeter and a 10 ohm shunt resistor.
Pattern Recognition
Notice how nicely the numbers are set up in JEE Mains questions! The parallel combination of 240Ω$240\Omega$ and 10Ω$10\Omega$ yields exactly 9.6Ω$9.6\Omega$, which beautifully combines with 150.4Ω$150.4\Omega$ to form a perfect integer sum of 160Ω$160\Omega$. This tells you that your intermediate steps are absolutely correct!
Chapter Mix
Class 12 Physics: Current Electricity: Measuring Devices
Keywords:#ammeter shunt current divider#galvanometer conversion ammeter#JEE Main 2025 Morning Q25#Ohm's law equivalent resistance#shunted ammeter#current divider rule#equivalent resistance circuit
More Current Electricity Previous-Year Questions — Page 2
Q36jee_main_2026_22_january_eveningPower Transmission and Efficiency
An electric power line having total resistance of 2 Ω$2 \Omega$, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ____.
A.96.9$96.9$
B.86.5$86.5$
C.100$100$
D.92.5$92.5$
Solution
Related Formula
Pout = V · I$$P_{\text{out}} = V \cdot I$$Ploss = I² R$$P_{\text{loss}} = I^2 R$$η = ( PoutPₙₑₜ) × 100%$$\eta = \left(\frac{P_{\text{out}}}{P_{\text{net}}}\right) \times 100\%$$
Core Logic
Calculating total current I$I$:
1000 = 250 × I I = 4 ~A$$1000 = 250 \times I \implies I = 4 \mathrm{~A}$$
Calculating power loss along the line Ploss$P_{\text{loss}}$:
The percentage efficiency of the transmission line is 96.9%$96.9\%$.
Pattern Recognition
Efficiency formula: η = PoutPout + I² R × 100%$\eta = \frac{P_{\text{out}}}{P_{\text{out}} + I^2 R} \times 100\%$.
Current I = 1000/250 = 4~A$I = 1000/250 = 4\mathrm{~A}$. Loss = 16 × 2 = 32~W$= 16 \times 2 = 32\mathrm{~W}$. η = 1000/1032 = 96.9%$\eta = 1000/1032 = 96.9\%$.
Chapter Mix
Class 12 Physics: Current Electricity
Q49jee_main_2026_22_january_eveningDrift Velocity and Electron Mobility
A cylindrical conductor of length 2m and area of cross-section 0.2 ~mm²$0.2 \mathrm{~mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is α × 10⁻³ ~m²/V⋯$\alpha \times 10^{-3} \mathrm{~m}^2/\mathrm{V}\cdot\mathrm{s}$. The value of α$\alpha$ is :
(electron concentration = 5 × 10²⁸/m³$5 \times 10^{28}/\mathrm{m}^3$ and electron charge = 1.6 × 10⁻¹⁹$1.6 \times 10^{-19}$ C)
Numerical Answer.Answer: 1 to 1
Solution
Related Formula
I = n e A vd$I = n e A v_d$
vd = μ E = μ (V)/(l)$$v_d = \mu E = \mu \frac{V}{l}$$μ = (I l)/(n e A V)$$\mu = \frac{I l}{n e A V}$$
Core Logic
Combining current density and mobility equations:
I = n e A (μ (V)/(l)) μ = (I · l)/(n · e · A · V)$$I = n e A \left(\mu \frac{V}{l}\right) \implies \mu = \frac{I \cdot l}{n \cdot e \cdot A \cdot V}$$
Substituting given values I = 1.6 ~A, l = 2 ~m, n = 5 × 10²⁸ /m³, e = 1.6 × 10⁻¹⁹ ~C, A = 0.2 × 10⁻⁶ ~m², V = 2 ~V$I = 1.6 \mathrm{~A}, l = 2 \mathrm{~m}, n = 5 \times 10^{28} /\mathrm{m}^3, e = 1.6 \times 10^{-19} \mathrm{~C}, A = 0.2 \times 10^{-6} \mathrm{~m}^2, V = 2 \mathrm{~V}$:
Mobility formula: μ = I l / (n e A V)$\mu = I l / (n e A V)$. Direct parameter plug-in yields α = 1$\alpha = 1$.
Chapter Mix
Class 12 Physics: Current Electricity
Q34jee_main_2026_23_january_morningResistance and Ohm's Law
A wire of uniform resistance λΩ/m$\lambda\Omega/m$ is bent into a circle of radius r and another piece of wire with length 2r is connected between points A and B (AOB) as shown in figure. The equivalent resistance between points A and B is ____ Ω$\Omega$.
A ring with nodes A and B connected across a diameter forming parallel branches.
The system represents three resistors connected in parallel between nodes A and B: the upper arc, the lower arc, and the straight diameter wire. Using R = λ L$R = \lambda L$ where λ$\lambda$ is resistance per unit length.
RAB = λ r ((6π)/(16 + 3π))$$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right)$$
Pattern Recognition
Sees: "Uniform resistance wire bent into shape" → immediately break the shape down into parallel/series segments defined purely by their arc/line lengths multiplied by the linear density λ$\lambda$.
To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ____
Note: The calculated exact percentage error is 1.16%, which does not match any of the provided options exactly. The closest official option provided was tracked as index (2) marking 1.65 by some keys, though technically a bonus question.
Chapter Mix
Class 12 Physics: Current Electricity
Class 11 Physics: Units and Measurements
Two resistors of 100 Ω$100 \, \Omega$ each are connected in series with a 9V battery. A voltmeter of 400Ω$400\Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ____ V.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
