Physics Current Electricity

A cell, shunted by a8 $Ω$ resistance, is balanced across a potentiometer wire of length 3m. The balancing length is 2m when cell is shunted by 4 $Ω$ resistance. The value of internal resistance of the cell will be__________ $Ω$ .

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R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_{0}} = \frac{V A}{ρ L}$ $\frac{}{_{}} \frac{}{}$ (1)

A Area of potentiometer wire

$ρ \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $Ω$ shunt is bal aced at 3m length

$V_{C D} = I R_{C D} = \frac{V}{ρ L} A * \frac{ρ C F}{A}$

$V_{C D} = \frac{V l}{L} = \frac{V * 3}{L}$

VCD = VAB = E – I1r

= $- \frac{E}{R + r} r = \frac{V}{L} * 3$ $\frac{}{} \frac{}{}$

$\Rightarrow E [1 - \frac{r}{8 + r}] = \frac{V}{L} * 3$ (2)

Case 2 When 4 $Ω$ shunt is balanced at 2m length

$V_{C F} = I R_{C F} = \frac{V}{ρ L} A * \frac{ρ C F}{A} = \frac{V * 2}{L}$

$V_{C F} = V_{A B} = E - I_{2} r$

= $E - \frac{E}{R + r} r$ $\frac{}{}$

= $E [1 - \frac{r}{4 + r}]$ $\frac{}{}$

$\frac{}{} \frac{}{}$ $\Rightarrow E (1 - \frac{r}{4 + r}) = \frac{V * 2}{L}$ (3)

$(2) \div (3)$

$\frac{1 - \frac{r}{8 + r}}{1 - \frac{r}{4 + r}} = \frac{3}{2}$

$\Rightarrow (\frac{8}{8 + r}) * (\frac{4 + 8}{4}) = \frac{3}{2}$

$\Rightarrow 4 (4 + r) = 3 (8 + r)$

$\Rightarrow 16 + 4 r = 24 + 3 r$

r = 24 – 16

r = 8 $Ω$

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

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

% of I which passes through galvanometer

$l_{g} = 10 %$ of I

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

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

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