Find the current through a resistance 2 ohm if the voltage across the resistance is 6 V

Step 1: Given: Resistance = 2\(\Omega\)

Voltage = 6 V

To find: Current, I =?

Step 2: Calculating current

From Ohm’s law, we know that,

V = IR

I = V/R

I = 6/2

I = 3 A

Therefore the value of the current through the resistance is 3 A.

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Find the current through a resistance 2 ohm if the voltage across the resistance is 6 V.

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Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
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Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

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Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

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Comments

SUBSCRIBE TO OUR NEWS LETTER

Find the current through a resistance 2 ohm if the voltage across the resistance is 6 V.

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CBSE CLASS XII Related Questions

1.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

3.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

5.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

3.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

5.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: