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a month ago

Overview 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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New answer posted

a month ago

Overview 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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New answer posted

a month ago

Overview 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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New answer posted

a month ago

Overview 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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New answer posted

a month ago

Overview 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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New answer posted

a month ago

Overview Class 12th

A resistor dissipates 192 J of energy in 1s when a current of 4A is passed4 through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in________J.

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Vishal Baghel

Contributor-Level 10

Using Heat equation : H = i²Rt

=>192 = (4)² R (1)

H = (8)² R (5)

=>H = 3840 J

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New answer posted

a month ago

Overview Class 12th

The equivalent resistance of the given circuit between the terminals A and B is:

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Vishal Baghel

Contributor-Level 10

Since $5 Ω$ is connected across conductor so we can remove it.

$R_{e q} = 1 Ω$

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New answer posted

2 months ago

Overview Class 12th

A square shaped wire with resistance of each side $3 Ω$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $Ω$ will be________.

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Raj Pandey

Contributor-Level 9

$R_{e q} = 3 Ω$

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New answer posted

2 months ago

Overview Class 12th

First, a set of n equal resistors of 10 $Ω$ each are connected in series to a battery of emf 20V and internal resistance 10 $Ω$ . A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increases 20 times, then the value of n is………

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Vishal Baghel

Contributor-Level 10

In series Req = nR = 10n

$\Rightarrow l s = \frac{20}{10 + 10 n} = \frac{2}{1 + n}$

In parallel $\Rightarrow R_{e q} = \frac{10}{n}$ $_{} \frac{}{}$

$\Rightarrow \frac{l_{ρ}}{l_{s}} = 20 = \frac{2 n / 1 + n}{2 / 1 + n}$

n = 20

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New answer posted

2 months ago

Overview Class 12th

Two cells of same emf but different internal resistance r₁ and r₂ are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, if:

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Raj Pandey

Contributor-Level 9

Given

$I = \frac{2 E}{R = r_{1} + r_{2}}$

E₁ = E₂ = E - (i)

Potential drop across second cell is

$V_{A} - E_{2} + I r_{2} = v_{B}$

According to question VA – VB = 0

E2 lr2 = 0

$\Rightarrow R = r_{2} - r_{1}$

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