Physics Current Electricity

A $72 Ω$ galvanometer is shunted by a resistance of 8 $Ω$ . The percentage of the total current which passes through the galvanometer is:

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Contributor-Level 10

$\Rightarrow I_{g} G = (l - l_{g}) 8 \Rightarrow l_{g} * 72 = (l - l_{g}) 8 \Rightarrow 9 l_{g} = l - l_{g}$

$10 l_{g} = l l_{g} = \frac{1}{10} l$

% of I which passes through galvanometer

$l_{g} = 10 %$ of I

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8 months ago

In a building there are 15 bulbs of $45 W, 15$ bulbs of $100 W, 15$ small fans of $10 W$ and 2 heaters of $1 k W$ . The voltage of electric main is $220 V$ . The minimum fuse capacity (rated value) of the building will be

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Contributor-Level 10

Total power is $(15 * 45) + (15 * 100) + (15 * 10) + (2 * 1000) = 4325 W$

So, current is $\frac{4325}{220} = 19.66 A$

Answer is $20 A m p$ .

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8 months ago

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r₁ and r₂ (r₁ > r₂). If the potential difference across the source of internal resistance r₁ is zero, then the value of R will be:

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Contributor-Level 10

V_AB = 0

$o r, ε - i r_{1} = 0$

$\Rightarrow ε - \frac{2 ε}{r_{1} + r_{2} + R} r_{1} = 0 \Rightarrow 1 = \frac{2 r_{1}}{r_{1} + r_{2} + R}$

$\Rightarrow R + r_{1} + r_{2} = 2 r_{1}$

$R = r_{1} - r_{2}$

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8 months ago

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be

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Payal Gupta

Contributor-Level 10

$H = l^{2} R t$

$% e r r o r i n H = \frac{Δ H}{H} * 100 = 2 (\frac{Δ l}{l}) * 100 + (\frac{Δ R}{R}) * 100 + (\frac{Δ t}{t}) * 100 = 2 * 2 + 1 + 3 = 8 %$

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8 months ago

A current of 15 mA flow in the circuit as shown in figure. The value of potential difference between the pointes A and B will be

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Payal Gupta

Contributor-Level 10

$R_{A B} = 5 + (10 / / 5) + 10 = 15 + \frac{10 * 5}{10 + 5} = 15 + \frac{10}{3} = \frac{55}{3} k Ω$

$V_{A B} = R_{A B} l = (\frac{55}{3} * 1 0^{3}) * (15 * 1 0^{- 3}) = 275 V$

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A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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Contributor-Level 10

$R_{2} = ρ \frac{2 l}{A / 2} = \frac{4 ρ l}{A} = 4 R_{1}$
$\frac{R_{2}}{R_{1}} = \frac{4}{1} \to 4 : 1$

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8 months ago

A 1m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

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Contributor-Level 10

R_w = 2R_y

$\Rightarrow ρ (\frac{2 x}{\frac{A}{2}}) = \frac{2 ρ (1 - x)}{A}$

$\frac{4 ρ x}{A} = \frac{2 ρ (1 - x)}{A}$

->4x = 2 – 2x Þ 6x = 2

x = $\frac{2}{6} = \frac{1}{3}$ $∴ \frac{L_{x}}{L_{y}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{3} * \frac{3}{2}$

$= \frac{1}{2}$

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8 months ago

The balancing length for a cell is $560 c m$ in a potentiometer experiment. When an external resistance of $10 Ω$ is connected in parallel to the cell, the balancing length changes by $60 c m$ . If the internal resistance of the cell is $\frac{N}{10} Ω$ , where $N$ is an integer then value of $N$ is

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Contributor-Level 10

Loss of energy $= \frac{1}{2} {C V}^{2} - \frac{1}{2} (2 C) * V_{Common}^{2}$ $\frac{}{}^{} \frac{}{}_{}^{}$

$\begin{array}{r} = \frac{1}{2} * 60 * 10^{- 12} * (20)^{2} - 60 * 10^{- 12} * (10)^{2} \\ = 60 * 10^{- 12} (200 - 100) \\ = 6000 * 10^{- 12} \\ = 6 n J \end{array}$

n=12

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8 months ago

As shown in the figure, in steady state, the charge stored in the capacitor is ____________* 10^-⁶C.

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Contributor-Level 10

$q = C V_{100 Ω}$

$= (1.1 * 1 0^{- 6}) (\frac{10}{R + r} R)$

$= 1.1 * 1 0^{- 6} (\frac{10}{110} * 100)$

$= 10 μ C$

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8 months ago

In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be__________cm.

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