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Wheatstone Bridge 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 month ago

Wheatstone Bridge Class 12th

A ball is projected with a velocity, $10 {m s}^{- 1}$ , at an angle of $60^{?}$ with the vertical direction, its speed at the highest pint of its trajectory will be :

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Raj Pandey

Contributor-Level 9

At the highest point only horizontal component of velocity remains $\Rightarrow u_{x} = u c o s ? θ$

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a month ago

Wheatstone Bridge Class 12th

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.

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alok kumar singh

Contributor-Level 10

Let V? = 10V, V? = xV, V? = 0V, and V? = yV.
Applying Kirchhoff's current law at node B:
(x - 10)/100 + (x - y)/15 + (x - 0)/10 = 0 ⇒ 53x - 20y = 30 . (1)
Applying Kirchhoff's current law at node D:
(y - 10)/60 + (y - x)/15 + (y - 0)/5 = 0 ⇒ 17y - 4x = 10 . (2)
Solving equations (1) and (2), we get:
x = 0.865 and y = 0.792
The current i? is:
i? = (x - y) / 15 = 4.87 mA

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New answer posted

a month ago

Wheatstone Bridge Class 12th

Four resistances of $15 Ω, 12 Ω, 4 Ω$ and $10 Ω$ respectively in cyclic order to form Whetstone's network. The resistance that is to be connected in parallel with the resistance of $10 Ω$ to balance the network is

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Raj Pandey

Contributor-Level 9

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New answer posted

2 months ago

Wheatstone Bridge Class 12th

In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance 30cm from the left side as shown at point D. If R is 5.6 $k Ω$ , then the value of unknown resistance 'S' will be ___________ $Ω .$

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Payal Gupta

Contributor-Level 10

$\frac{30}{70} = \frac{S}{5.6 k Ω} \Rightarrow S = \frac{3}{7} * 5.6 k Ω = 3 * 0.8 k Ω$

= 2.4k $Ω$ = 2400 $Ω$

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New answer posted

2 months ago

Wheatstone Bridge Physics NCERT Exemplar Solutions Class 12th Chapter Three

Four resistance $40 Ω, 60 Ω, 90 Ω$ , and $110 Ω$ make the arms of a quadrilateral ABCD. Across $A C$ is a battery of emf $40 V$ and internal resistance negligible. The potential difference across $B D$ in $V$ is

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alok kumar singh

Contributor-Level 10

10553.33

Sol. $\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

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