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3 months ago

Overview Class 12th

In connection with the circuit drawn below, the value of current flowing through 2 $k Ω$ resistor is________ * 10^-4A

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alok kumar singh

Contributor-Level 10

Potential difference across 2k $Ω$ is 5V, thus current through it,

$i = \frac{5}{2 * 1 0^{3}} = 25 * 1 0^{- 4} A .$

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3 months ago

Overview Class 12th

A wire of $1 Ω$ has a length of 1m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is:

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Vishal Baghel

Contributor-Level 10

$R_{i} = \frac{ρ l}{A}$

$R_{f} = \frac{ρ (1.25 l)}{(A / 1.25)} = {(1.25)}^{2} * \frac{ρ l}{A}$

$\Rightarrow R_{f} = 1.5625 * R_{i}$

$\Rightarrow \frac{R_{f} - R_{l}}{R_{l}} * 100 = 56.25 %$

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New answer posted

3 months ago

Overview Class 12th

In an electrical circuit, a battery is connected to pass 20 C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15V. The work done by the battery is--------- J.

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Payal Gupta

Contributor-Level 10

$W = q Δ V = 20 * 15 = 300 J$

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3 months ago

Overview Class 12th

Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is

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Payal Gupta

Contributor-Level 10

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Overview Class 12th

A cylindrical wire of radius 0.5mm and conductivity 5 * 10⁷ S/m is subjected to an electric field of 10m V/m. The expected value of current in the wire will be x³πmA. The value of x is………………….

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Vishal Baghel

Contributor-Level 10

radius r = 0.5 mm

$σ = 5 * 1 0^{7} s / m$

$\vec{E} = 10 * 1 0^{- 3} V / m = 1 0^{- 2} V / m$

As we know

$J = σ E$

$J = \frac{i}{A} = σ E$

$i = σ E A = 5 * 1 0^{7} * 1 0^{- 2} * π * {(0.5 * 1 0^{- 3})}^{2} = 1.25 π * 1 0^{- 1} A = 125 π m A = x^{3} π m A$

$∴$ x = 5

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3 months ago

Overview Class 12th

In the given circuit of potentiometer, the potential difference E across AB (10m length) is larger than E₁ and E₂ as well. For key K₁(closed), the jokey is adjusted to touch the wire at point J₁ so that there is no deflection in the galvanometer. Now the first battery (E₁) is replaced by second battery (E₂) for working by making K₁ open and K₂ closed. The galvanometer gives then null deflection at J₂. The value of $\frac{E_{1}}{E_{2}} i s \frac{a}{b},$ where a = …………….

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Vishal Baghel

Contributor-Level 10

$\frac{E_{1}}{E_{2}} = \frac{380}{760} = \frac{1}{2}$

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3 months ago

Overview Class 12th

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance $2 Ω$ each and leaves by the corner R. The currents i₁ in ampere is………

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Payal Gupta

Contributor-Level 10

i₂ = 2i₁

i₁ + i₂ = 6

i₁ = 2A.

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Overview Class 12th

Two heaters $A$ and $B$ have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:

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alok kumar singh

Contributor-Level 10

Power Consumed $= P = \frac{V^{2}}{R}$

$\frac{P_{A}}{P_{B}} = \frac{R_{B}}{R_{A}}$

$R_{A} = 2 R_{B}$

For Series Combination

$P_{S} = \frac{V^{2}}{3 R_{B}}$

For Parallel Combination

$P_{P} = \frac{3 V^{2}}{2 R_{B}}$

$\frac{P_{S}}{P_{P}} = \frac{2}{9}$

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3 months ago

Overview Class 12th

Choose the correct circuit which can achieve the bridge balance.

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alok kumar singh

Contributor-Level 10

In option (1),

$\frac{10}{15} = \frac{10}{5 + R_{D}}$

The diode can conduct and have resistance $R_{D} = 10 Ω$ because diode have dynamic resistance. In that case bridge will be balanced.

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3 months ago

Overview Class 12th

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{2}$ is 40cm. Now as unknown resistance x is connected in series with P and new balancing length is found to be 80cm measured from the same end. Then the value of x will be _________ $Ω$ .

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Payal Gupta

Contributor-Level 10

Initially, $\frac{P}{Q} = \frac{40}{60} = \frac{2}{3}$ - (1)

Finally, $\frac{P + x}{Q} = \frac{80}{20} = \frac{4}{1}$

(2) (1)

$\frac{P + x}{P} = \frac{4 * 3}{2} = 6$

$\Rightarrow 1 + \frac{x}{P} = 6$

$\frac{x}{P} = 5$

x = 5 P = 5 * 4 = 20 $Ω$

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