In this figure the resistance of the coil of galvanometer G is 2Ω. The emf of the cell is 4 V. The ratio of potential difference across C1 and C2 is
- \(\frac{4}{5}\)
- 1
- \(\frac{5}{4}\)
- \(\frac{3}{4}\)
The Correct Option is A
Solution and Explanation
Step 1: Apply Kirchhoff’s laws.- Using the given circuit, calculate the currents and potential drops across C1 and C2.- The ratio is determined as:
$\frac{V_{C1}}{V_{C2}} = \frac{4}{5}$.
Final Answer: The ratio of potential differences is $\frac{4}{5}$
Top Questions on Resistance
- A carbon resistor has a tolerance of 20\%. As per the colour codes of resistors, the last band in that resistor is
- KEAM - 2025
- Physics
- Resistance
- A current of 2 A flows through a conductor with a resistance of 5 \(\Omega\). Calculate the potential difference across the conductor.
- MHT CET - 2025
- Physics
- Resistance
- A coil of resistance 10 \( \Omega \) is connected to a battery of 12 V. If the current flowing through the coil is 2 A, what is the power dissipated in the coil?
- BITSAT - 2025
- Physics
- Resistance
- What is the resistance of a conductor if the potential difference across it is 12 V and the current flowing through it is 3 A?
- VITEEE - 2025
- Physics
- Resistance
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:
- JEE Main - 2025
- Physics
- Resistance
Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: