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2 months ago

Overview Class 12th

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $Ω$ is used to measure the potential difference across 500 $Ω$ resister, the reading of the voltmeter will be_________ V.

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Vishal Baghel

Contributor-Level 10

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0 $\Rightarrow I = \frac{20}{5000} A$

Reading of voltmeter = 2000I = 2000 * $\frac{20}{5000} = 8 v o l t$

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New answer posted

2 months ago

Overview Class 12th

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω .$ The value of internal resistance of each cell is

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$\frac{2 V}{2 + 2 r} = \frac{V}{2 + \frac{r}{2}}$

$\Rightarrow 2 + 2 r = 4 + r \Rightarrow r = 2 Ω$

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New answer posted

2 months ago

Overview Class 12th

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $Ω$ is used to measure the potential difference across 500 $Ω$ resister, the reading of the voltmeter will be_________ V.

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0 $\Rightarrow I = \frac{20}{5000} A$

Reading of voltmeter = 2000I = 2000 * $\frac{20}{5000} = 8 v o l t$

0 0

New answer posted

2 months ago

Overview Class 12th

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

According to question, we can write

$\frac{2 V}{2 + 2 r} = \frac{V}{2 + \frac{r}{2}}$

$\Rightarrow 2 + 2 r = 4 + r \Rightarrow r = 2 Ω$

0 0

New answer posted

2 months ago

Overview Class 12th

A simple $L R$ circuit is connected to a battery at $t = 0$ . The time instant at which rate of energy storage in inductor is half of power delivered by battery

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alok kumar singh

Contributor-Level 10

$i = \frac{v_{0}}{R} (1 - e^{\frac{t R}{L}})$

$U = \frac{1}{2} L i^{2}$

$\frac{d U}{d t} = L i \frac{d i}{d t}$

$2 L * \frac{R}{L} * \frac{v_{0}^{2}}{R^{2}} (1 - e^{\frac{t R}{L}}) e^{\frac{t R}{L}} = \frac{v_{0}^{2}}{R} (1 - e^{\frac{t R}{L}})$

$t = \frac{L}{R} I n 2$

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New question posted

2 months ago

Overview Class 12th

In the circuit in figure we have $ε_{1} = ε_{2}$ = 100V, R₁ = 20 $Ω$ , R₂ = 10 $Ω$ , R₃ = 40 $Ω$ and R₄ = 30 $Ω$ . Find the reading of the ammeter (in A). Disregard the resistance of the battery and the ammeter.

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New answer posted

2 months ago

Overview Class 12th

In half deflection method, the resistance of resistance box in series with the galvanometer is 10,000 $Ω$ . When the galvanometer is shunted with 10 $Ω$ resistance, its reading drops to half the initial value. The resistance of galvanometer is:-

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Vishal Baghel

Contributor-Level 10

$i = \frac{v}{R + G}$

$\frac{i}{2} = \frac{v}{\frac{S G}{S + G} + R} * \frac{S}{S + G}$

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

$\frac{S R}{R - S} = G$

R > > S

$\Rightarrow G \approx S$

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New answer posted

2 months ago

Overview Class 12th

Write the value of current i₃ (in mA) in the circuit shown in the figure.

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alok kumar singh

Contributor-Level 10

$5 * 1 0^{3} I_{1} = 60$

$\Rightarrow I_{1} = \frac{60}{5 * 1 0^{3}} = 12 m A$

Potential difference across

$10 k Ω = 240 - 60 = 180 V$

$I_{2} = I_{3} - I_{1} = 6 m A$

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New answer posted

2 months ago

Overview Class 12th

In the circuit shown swith S is connected to position 2 for a long time and then joined to position 1. The total heat produced in resistance R₁ is:

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alok kumar singh

Contributor-Level 10

$I = \frac{E}{R_{2}}$

$U_{I n d u c t o r} = H e a t l o s s i n R_{1} = \frac{1}{2} L {(\frac{E}{R_{2}})}^{2}$

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New answer posted

2 months ago

Overview Class 12th

The circuit shown in the figure is in steady state for a long time. The connection to battery is suddenly broken (switch S is opened up). What is the charge (in $μ C$ ) on the capacitor after 0.001 sec.?