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Physics Current Electricity

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Physics Current Electricity Class 12th

A current through a wire-depends on time as

i = α₀t + βt²

Where α₀ = 20A/s and β = 8As^-². Find the charge crossed through a section of the wire in 15s.

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Vishal Baghel

Contributor-Level 10

$\frac{d q}{d t} = t = 20 t + 8 t^{2}$

$\Rightarrow \int_{0}^{q} d q = \int_{0}^{15} (20 t + 8 t^{2}) d t$

$\Rightarrow q = 20 * \frac{1 5^{2}}{2} + 8 . \frac{1 5^{3}}{3} = 11250 C$

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Physics Current Electricity Class 12th

As per the given circuit, the value of current through the battery will be__________A.

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Raj Pandey

Contributor-Level 9

$\begin{array}{l} D_{1} \\ D_{3} \end{array}}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I = \frac{v}{R} = \frac{10}{10} = 1 A m p$

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Physics Current Electricity Class 12th

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is:

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Raj Pandey

Contributor-Level 9

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} * 100 = (\frac{d l}{l} - \frac{d A}{A}) * 100$

= 0.4 - (-0.4)

= 0.8 %

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Physics Current Electricity 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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a year ago

Physics Current Electricity 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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a year ago

Physics Current Electricity Class 12th

Current measured by the ammeter (A) in the reported circuit when no current flows through $10 Ω$ resistance, will be A.

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Raj Pandey

Contributor-Level 9

Using whetstone

$\frac{R}{3} = \frac{4}{6} \Rightarrow R = 2 Ω$

$R_{e q u} = \frac{54}{15} Ω$

$i = \frac{36}{\frac{54}{15}} = 10 A$

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a year ago

Physics Current Electricity Class 12th

A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 $Ω$ . The galvanometer (G) shows null deflection when tapping key is at 43cm mark from end A. If the end correction for end A is 2cm. Then the determined value of R will be ________ $Ω$ .

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Raj Pandey

Contributor-Level 9

At null point:

$\frac{15}{43 + 2} = \frac{R}{100 - 43}$

$\Rightarrow R = 19 Ω$

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New answer posted

a year ago

Physics Current Electricity 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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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 * $\frac{20}{5000} = 8 v o l t$

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New answer posted

a year ago

Physics Current Electricity 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

a year ago

Physics Current Electricity 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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