Ask & Answer Question

Electric Current Physics Current Electricity Physics Class 12th

The value of current in the $6 Ω$ resistance is:

Option 1 -

Option 2 -

10A

Option 3 -

Option 4 -

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

Method-I : Using Kirchhoff’s Law for loop $\vec{B D A B B},$ we can write

$- 140 + 20 l + 6 l_{1} = 0 \Rightarrow 10 l + 3 l_{1} = 70 . . . . . . . (i)$

Using Kirchhoff’s Law for loop $\vec{A C B B A},$

we can write

$5 (l - l_{1}) + 90 - 6 l_{1} = 0$

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25 l_{1} = 250 \Rightarrow l_{1} = \frac{250}{25} = 10 A$

Method-II : Using concept of equivalent cell, we can write

$V_{A B} = \frac{\frac{140}{20} + \frac{0}{6} + \frac{90}{5}}{\frac{1}{20} + \frac{1}{6} + \frac{1}{5}} = \frac{2100 + 0 + 5400}{15 + 50 + 60}$

$= \frac{7500}{125} = 60 V o l t$

$\Rightarrow l_{1} = \frac{60}{6} = 10 A$

Method-I : Using Kirchhoff’s Law for loop <math> <mrow> <mover accent="true"> <mrow> <mi>B</mi> <mi>D</mi> <mi>A</mi> <mi>B</mi> <mi>B</mi> </mrow> <mo stretchy="true">→</mo> </mover> <mo>,</mo> </mrow> </math>  we can write <math> <mrow> <mo>−</mo> <mn>1</mn> <mn>4</mn> <mn>0</mn> <mo>+</mo> <mn>2</mn> <mn>0</mn> <mi>l</mi> <mo>+</mo> <mn>6</mn> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mn>0</mn> <mo>⇒</mo> <mn>1</mn> <mn>0</mn> <mi>l</mi> <mo>+</mo> <mn>3</mn> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mn>7</mn> <mn>0</mn> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mrow> <mo>(</mo> <mrow> <mi>i</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math>              Using Kirchhoff’s Law for loop <math> <mrow> <mover accent="true"> <mrow> <mi>A</mi> <mi>C</mi> <mi>B</mi> <mi>B</mi> <mi>A</mi> </mrow> <mo stretchy="true">→</mo> </mover> <mo>,</mo> </mrow> </math>  we can write <math> <mrow> <mn>5</mn> <mrow> <mo>(</mo> <mrow> <mi>l</mi> <mo>−</mo> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <mn>9</mn> <mn>0</mn> <mo>−</mo> <mn>6</mn> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math>              ->-5l + 11l1 = 90 .(ii)Adding equation (1) with twice the equation (2), we have <math> <mrow> <mn>2</mn> <mn>5</mn> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mn>2</mn> <mn>5</mn> <mn>0</mn> <mo>⇒</mo> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>0</mn> <mi>A</mi> </mrow> </math>              Method-II : Using concept of equivalent cell, we can write <math> <mrow> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>A</mi> <mi>B</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>4</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mn>0</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>0</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>9</mn> <mn>0</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> <mn>0</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>5</mn> <mn>4</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>5</mn> <mo>+</mo> <mn>5</mn> <mn>0</mn> <mo>+</mo> <mn>6</mn> <mn>0</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>7</mn> <mn>5</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>6</mn> <mn>0</mn> <mi>V</mi> <mi>o</mi> <mi>l</mi> <mi>t</mi> </mrow> </math> <math> <mrow> <mo>⇒</mo> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mn>0</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>0</mn> <mi>A</mi> </mrow> </math>              <div class="photo-widget-full"><div class="figure"><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754566400phpU1Kkoi.jpeg" alt="" width="400" height="225" loading="lazy"></picture></div></div>

0 Upvotes 0 Downvotes

Similar Questions for you

A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

alok kumar singh

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 \times \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

$R_{eq} = R_{S} + R_{P}$

$= 52 Ω$

A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 \times 1 0^{9}$ SI units)

alok kumar singh

For uniformly charged spherical shell,

$V = \frac{k q}{R} (For r \leq R)$

$∴ V_{C} = V_{P}$

$V_{C} - V_{P} = Zero$

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

alok kumar singh

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

x = 4

The terminal voltage of the battery, whose emf is 10 V and internal resistance $1 Ω$ , when connected through an external resistance of $4 Ω$ as shown in the figure is:

alok kumar singh

$Current in circuit i = \frac{10}{4 + 1} = 2 A$

$Terminal voltage = E - i R$

$= 10 - 2 \times 1 = 8 V$

A current of 5A is passing through a non-linear magnesium wire of cross – section 0.04m². At every point the direction of current density is at an angle of 60° ;with unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is: (Resistivity of magnesium $ρ = 44 \times 1 0^{- 8} Ω m)$

alok kumar singh

$l = J A c o s θ \Rightarrow 5 = J \times 0.04 \times c o s 60 ° \Rightarrow J = \frac{5}{0.02} = 250 A / m^{2}$

$J = \frac{E}{ρ} \Rightarrow E = ρ J = 44 \times 1 0^{- 8} \times 250 = 11 \times 1 0^{- 5} V / m$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

The value of current in the $6 Ω$ resistance is:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

The value of current in the 6 Ω resistance is:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

The value of current in the $6 Ω$ resistance is: