Physics Current Electricity

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be:

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Contributor-Level 10

Current at full deflection,

$l_{m a x} = \frac{50}{2} = 25 m A$

$R = \frac{V}{l_{m a x}} = \frac{50}{25} = 2 K Ω$

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2 months ago

The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:

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Contributor-Level 10

$R = 75 * 1 0^{2} \pm 5 %$

$\Rightarrow R = 7500 \pm 375 Ω$

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2 months ago

Five identical cells each of internal resistance $1 Ω$ and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?

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Contributor-Level 10

$l_{1} = \frac{25}{5 + R}$

$l_{2} = \frac{5}{R + \frac{1}{5}}$

$\begin{array}{l} ? l_{1} = l_{2} \Rightarrow 4 R = 4 \\ \Rightarrow R = 1 Ω \end{array}$

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2 months ago

What equal length of an iron wire and a copper-nickel alloy wire, each of 2mm diameter connected parallel to give an equivalent resistance of 3 $Ω$ ?

(Given resistivities of iron and copper-nickel alloy wire are $12 μ Ω$ cm and 51 $μ Ω$ cm respectively)

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Contributor-Level 10

$\frac{R_{1} R_{2}}{R_{1} + R_{2}} = 3$

$\Rightarrow \frac{(ρ_{1} \frac{l}{A}) (ρ_{2} \frac{l}{A})}{(\frac{ρ_{1} l}{A}) + (\frac{ρ_{2} l}{A})} = 3$

$= \frac{3 * [\frac{π}{4} * {(2 * 1 0^{- 3})}^{2}] [12 + 51] * 1 0^{- 6} * 1 0^{- 2}}{12 * 1 0^{- 6} * 1 0^{- 2} * 51 * 1 0^{- 6} * 1 0^{- 2}}$

$= 97$

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2 months ago

If you are provided a set of resistance 2 $Ω$ , 4 $Ω$ , 6 $Ω$ and 8 $Ω$ . Connected these resistances so as to obtain an equivalent resistance of $\frac{46}{3} Ω .$

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Contributor-Level 10

$R_{e q} = \frac{2 * 4}{2 + 4} + 6 + 8 = \frac{46}{3} Ω$

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2 months ago

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200V supply. Calculate resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500W.

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Contributor-Level 10

As power delivered by the bulb is 500 W, potential difference across it should be 00 V. Thus,

$R = R_{b u l b} = \frac{10 0^{2}}{500} = 20 Ω$

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2 months ago

In the given figure, the emf of the cell is 2.2V and if internal resistance is 0.6 $Ω$ . Calculate the power dissipated in the whole circuit:

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Contributor-Level 10

$R_{e q} = 0.6 Ω + \frac{\frac{4 * 12}{4 + 12} * \frac{6 * 8}{6 + 8}}{\frac{4 * 12}{4 + 12} + \frac{6 * 8}{6 + 8}} Ω$

$= 0.6 Ω + \frac{3 * \frac{48}{14}}{3 + \frac{48}{14}} Ω$

$= 0.6 Ω + \frac{(144 / 14) Ω}{(\frac{90}{14})}$

$\begin{array}{l} = 0.6 Ω + 1.6 Ω \\ = 2.2 Ω \end{array}$

p = $\frac{v^{2}}{R_{e q}} = \frac{{(2.2 V)}^{2}}{2.2 Ω} = 2.2 W$

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2 months ago

A teacher in his physics laboratory allotted an experiment ot determine the resistance (G) of a galvanometer. Students took the observations for $\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of G?

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Contributor-Level 10

In galvanometer

$\Rightarrow (l - l_{g}) S = l_{g} G$

$\frac{I_{g}}{l} = \frac{S}{S + G}$

$\Rightarrow \frac{1}{3} = \frac{S}{S + G} \Rightarrow S + G = 3 S \Rightarrow G = 2 S$

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2 months ago

The length of a give cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be…………….%.

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Contributor-Level 10

volume of wire will remains constant

$π r_{1}^{2} l_{1} = π r_{2}^{2} l_{2}$

$\Rightarrow l_{2} = 2 l_{1} (G i v e n)$

$r_{2} = \frac{r_{1}}{\sqrt[]{2}}$

$R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}} = \frac{ρ 2 l_{1}}{π {(\frac{r_{1}}{\sqrt[]{2}})}^{2}}$

$= \frac{R_{2} - R_{1}}{R_{1}} * 100 = \frac{4 R_{1} - R_{1}}{R_{1}} * 100 = \frac{3 R_{1}}{R_{1}} * 100 = 300 %$

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2 months ago

If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:

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