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Physics Current Electricity

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2 months ago

Physics Current Electricity Class 12th

As shown in the figure, a potentiometer wire of resistance $20 Ω$ and length 300cm is connected with resistance box (R.B) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is __________ $Ω .$

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Vishal Baghel

Contributor-Level 10

$R_{A B} = 20 Ω$

L = 300 cm

Null point is at 60 cm from A, so

$20 m V = \frac{4}{R + 20} * 20 * (\frac{60}{300}) * 1 0^{3}$

$\Rightarrow 20 = \frac{16}{R + 20} * 1 0^{3}$

$R = \frac{15600}{20} = 780 Ω$

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2 months ago

Physics Current Electricity Class 12th

The total current supplied to the circuit is shown in figure by the 5 V battery is…………….A.

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Payal Gupta

Contributor-Level 10

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2 months ago

Physics Current Electricity Class 12th

A resistor develops 300 J of thermal energy in 15s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10s is…………….J.

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Payal Gupta

Contributor-Level 10

$H = I^{2} R t = 2^{2} * R * 15$

$\Rightarrow 300 = 60 R$

$R = 5 Ω$

Now for 3A, time = 10 sec

$H' = l^{2} R t = 3^{2} * 5 * 10 = 450 J$

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2 months ago

Physics Current Electricity Class 12th

A potentiometer wire of length 10m and resistance 20 $Ω$ is connected in series with a 25 V battery and an external resistance $30 Ω$ . A cell of emf E in secondary circuit is balanced by 250cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10} .$ The value of x is ______.

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Payal Gupta

Contributor-Level 10

AB = 10 m

$R_{A B} = 20 Ω$

$L_{A C} = 250 c m$

= 2.5 m

$l = \frac{25}{20 + 30}$

$E = \frac{x}{10} = \frac{25}{10}$

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2 months ago

Physics Current Electricity Class 12th

What will be the most suitable combination of three resistors A = $2 Ω$ , B = $4 Ω$ , C = $6 Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination

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Payal Gupta

Contributor-Level 10

Given

$\begin{array}{l} A = 2 Ω \\ B = 4 Ω \\ C = 6 Ω \\ R_{e q} = \frac{2 * 4}{2 + 4} + 6 \end{array}$

$R_{e q} = \frac{22}{3} Ω$

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2 months ago

Physics Current Electricity Class 12th

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20Ω . A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

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Vishal Baghel

Contributor-Level 10

$E_{e q} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}}} = \frac{\frac{F}{r} + \frac{E}{r}}{\frac{1}{r} + \frac{1}{r}} + \frac{\frac{2 E}{r}}{\frac{2}{r}} = \frac{\frac{2 E}{r}}{\frac{2}{r}} = E$

$r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{r r}{2 r} = \frac{r}{2}$

E_eq = E = 1.5 v

r_eq = $\frac{r}{2}$

$v_{A} = l * 10 = 1.2 = \frac{E}{10 + r / 2} * 10$

$\Rightarrow 1.2 = \frac{10 E}{10 + r / 2}$

$\Rightarrow 1.2 = \frac{10 * 1.5}{10 + r / 2}$

$\Rightarrow$ 12 + 0.6r = 15

0.6r = 3

^{$r = \frac{30}{6} = 5 Ω$}

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New answer posted

2 months 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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Payal Gupta

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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2 months 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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Payal Gupta

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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2 months ago

Physics Current Electricity Class 12th

A capacitor is discharging through a resistor R. Consider in time t₁ the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored in the capacitor reduces to half of its initial value and in time t₂ the charge stored reduces to one eight of its initial value. The ratio t₁/t₂ will be

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Payal Gupta

Contributor-Level 10

According to Law of discharging of capacitor, we can write

$Q = Q_{0} e^{- \frac{t}{τ}} \Rightarrow \frac{Q_{2}}{8} = Q_{0} e^{- \frac{t_{2}}{τ}} \Rightarrow t_{2} 3 τ l n (2), a n d$

$U = \frac{Q^{2}}{2 C} = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow \frac{1}{2} (\frac{Q_{0}^{2}}{2 C}) = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow t_{1} = \frac{τ}{2} l n (2)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{6}$

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2 months ago

Physics Current Electricity Current Electricity Analysis

A zener of breakdown voltage V_Z = 8V and maximum zener current, I_ZM = 10mA is subjected to an input voltage V_i = 10V with series resistance R = 100 $Ω$ In the given circuit R_L represents the variable load resistance. The ratio maximum and minimum value of R_L is__________.

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alok kumar singh

Contributor-Level 10

Across zener diode & RL

8 = I_LR_L

8 = $(20 - 10) R_{L (m a x)}$

$R_{L m a x} = \frac{8}{10} k Ω$

At max current in loop (1)

10 = i_max R + 8

i_max= $\frac{2}{R} = \frac{2}{100} = 0.02 A$

= 20 mA

R_L (max) = $\frac{8}{10} k Ω$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20} k Ω$

$\frac{R_{L m a x}}{R_{L m i n}} = 2$

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