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Physics Current Electricity 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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Physics Current Electricity 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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Physics Current Electricity Class 12th

A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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alok kumar singh

Contributor-Level 10

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 * \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

$R_{eq} = R_{S} + R_{P}$

$= 52 Ω$

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Physics Current Electricity Class 12th

A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 * 1 0^{9}$ SI units)

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alok kumar singh

Contributor-Level 10

For uniformly charged spherical shell,

$V = \frac{k q}{R} (For r \leq R)$

$∴ V_{C} = V_{P}$

$V_{C} - V_{P} = Zero$

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Physics Current Electricity Class 12th

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

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alok kumar singh

Contributor-Level 10

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

x = 4

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Physics Current Electricity 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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Physics Current Electricity Class 12th

The current I in the above given circuit will be

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Payal Gupta

Contributor-Level 10

$R = \frac{8}{2} Ω = 4 Ω$ (wheat stone bridge)

$I = \frac{V}{R} = \frac{40}{4} = 10 A$

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Physics Current Electricity Class 12th

The terminal voltage of the battery, whose emf is 10 V and internal resistance $1 Ω$ , when connected through an external resistance of $4 Ω$ as shown in the figure is:

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alok kumar singh

Contributor-Level 10

$Current in circuit i = \frac{10}{4 + 1} = 2 A$

$Terminal voltage = E - i R$

$= 10 - 2 * 1 = 8 V$

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Physics Current Electricity Class 12th

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5 Ω$ . However, it provides a potential difference of 1V across a load resistance of $2 Ω .$ The emf of the cell is given by $\frac{x}{10} V .$ Then the value of x is_________.

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Vishal Baghel

Contributor-Level 10

$i = \frac{ε}{R + r}$

$v_{R} = \frac{ε R}{R + r}$

$\frac{1.25 = \frac{ε (5)}{5 + r}}{1 = \frac{ε (2)}{2 + r}}$

r = 1

Using above equ : $ε = \frac{3}{2} = \frac{15}{10} V$

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Physics Current Electricity Class 12th

A Copper (Cu) rod of length 25cm and cross – sectional area 3 mm² is joined with a similar aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.

(Take Resistivity of Copper = 1.7 * 10^-⁸ $Ω m$

Resistivity of Aluminium = 2.6 * 10^-⁸ $Ω m$ )

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Vishal Baghel

Contributor-Level 10

$R_{1} = ?_{1} \frac{l}{A} R_{2} = ?_{2} \frac{l}{A}$

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{A}{l} [\frac{1}{?_{1}} + \frac{1}{?_{2}}]$

$= \frac{3 * 1 0^{? 6}}{\frac{1}{4}} [\frac{1}{1.7 * 1 0^{? 8}} + \frac{1}{2.6 * 1 0^{? 8}}]$

$R = 0.858 * 1 0^{? 3} ?$

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