Two bulbs rated (P1, V) and (P2, V) are connected (i) in series and (ii) in parallel across supply V. Determine power in the two combinations P1 and P2.

Electric power can be defined as the rate at which work is done or energy is converted into an electric circuit. In other words, it is the measurement energy used in a span of time.

Step 1:Formula

Electric Power formula is given by:

P = VI

Here (in the circuit),

P = Power
V = Potential Difference
I = Electric Current

Power is often also denoted as,

⇒ P = I²R
⇒ \(P = {{V^2} \over R}\)
\(\therefore\) Expressions above are determined using Ohm’s Law, wherein, Voltage, current, and resistance are interrelated by the relation:

V = IR

Here (in the circuit),

R = Resistance
V = Potential Difference
I = Electric Current

Step 2: Solution

(i) Two bulbs of (P1, V) and (P2, V) that are connected in series:

The flow of current in both bulbs is the same. Assuming R1 and R2 are both resistances of the bulbs, then the effective resistance circuit:

⇒ R = R1 + R2

Now, after using the relation, we can obtain:

P=V²/R

And, R = V²/P

Therefore, V²/P = V1²/ P1 + V2²/P2

As is given, V₁ = V₂ = V.

Therefore,

⇒ 1/P = 1/P₁ + 1/P₂

Hence, effective power of two bulbs connected in series is 1/ P = 1 / P₁ + 1 / P₂

(ii) Two bulbs of (P1, V) and (P2, V) when connected in parallel:

⇒ 1/R = 1/R₁ + 1/R₂

We know that

P = V²/R

1/R = V²/P

As is given, V1 = V2

Thus, P/V2 = P1/V² + P2/V²

Hence, the effective power of two bulbs connected in parallel is P = P₁ + P₂.

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A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

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View Solution

Two bulbs rated (P1, V) and (P2, V) are connected (i) in series and (ii) in parallel across supply V. Determine power in the two combinations P1 and P2.

Step 1:Formula

Step 2: Solution

CBSE CLASS XII Related Questions

1.
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:

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

4.
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

6.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

Similar Physics Concepts

Comments

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Two bulbs rated (P1, V) and (P2, V) are connected (i) in series and (ii) in parallel across supply V. Determine power in the two combinations P1 and P2.

Step 1:Formula

Step 2: Solution

CBSE CLASS XII Related Questions

1.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:

2.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

4.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

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

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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:

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

4.
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

6.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.