Question

. Find the quantity of current in the circuit if three resistance 2Ω, 3Ω,5 Ω are connected in series and voltage is 1.5V

Answer 1

Solution :

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(1,5)\qbezier(1,1)(1,1)(3,1)\qbezier(3,0.5)(3,0.5)(3,1.5)\qbezier(3.2,0.7)(3.2,0.7)(3.2,1.3)\qbezier(3.4,0.5)(3.4,0.5)(3.4,1.5)\qbezier(3.6,0.7)(3.6,0.7)(3.6,1.3)\qbezier(3.8,0.5)(3.8,0.5)(3.8,1.5)\qbezier(4,0.7)(4,0.7)(4,1.3)\qbezier(4,1)(4,1)(11,1)\qbezier(11,1)(11,1)(11,5.1)\qbezier(1,5)(1,5)(1.6,5)\qbezier(1.6,5)(1.6,5)(1.8,5.4)\qbezier(1.8,5.4)(1.8,5.4)(2,4.8)\qbezier(2,4.8)(2,4.8)(2.2,5.4)\qbezier(2.2,5.4)(2.2,5.4)(2.4,4.8)\qbezier(2.4,4.8)(2.4,4.8)(2.6,5.4)\qbezier(2.6,5.4)(2.6,5.4)(2.8,4.8)\qbezier(2.8,4.8)(2.8,4.8)(2.9,5.1)\qbezier(2.9,5.1)(2.9,5.1)(3.5,5.1)\qbezier(3.5,5.1)(3.5,5.1)(3.7,5.4)\qbezier(3.7,5.4)(3.7,5.4)(3.9,4.8)\qbezier(3.9,4.8)(3.9,4.8)(4.1,5.4)\qbezier(4.1,5.4)(4.1,5.4)(4.3,4.8)\qbezier(4.3,4.8)(4.3,4.8)(4.5,5.4)\qbezier(4.5,5.4)(4.5,5.4)(4.7,4.8)\qbezier(4.7,4.8)(4.7,4.8)(4.8,5.1)\qbezier(4.8,5.1)(4.8,5.1)(5.5,5.1)\qbezier(5.5,5.1)(5.5,5.1)(5.7,5.4)\qbezier(5.7,5.4)(5.7,5.4)(5.9,4.8)\qbezier(5.9,4.8)(5.9,4.8)(6.1,5.4)\qbezier(6.1,5.4)(6.1,5.4)(6.3,4.8)\qbezier(6.3,4.8)(6.3,4.8)(6.5,5.4)\qbezier(6.5,5.4)(6.5,5.4)(6.7,4.8)\qbezier(6.7,4.8)(6.7,4.8)(6.8,5.1)\qbezier(6.8,5.1)(6.8,5.1)(11,5.1)\put(3,1){\vector(-1,0){1}}\put(1.9,5.8){ \sf 2 \Ohm }\put(3.9,5.8){ \sf 3 \Ohm }\put(5.9,5.8){ \sf 5 \Ohm }\put(2.1,0.5){\sf 1.5 A} \put(3.4,-0.3){\sf 1.5 \ V}\end{picture}$

Equivalent Resistance : 10Ω

V = 1.5 V

I = V/R = 1.5/10 = 0.15 A .

Answer : Current in the circuit is 0.15 A

_________________________________________

Answer 2

Answer:-

• R1=2Ω
• R2=3Ω
• R3=5Ω
• Potential difference=V=1.5V
• Current=I=?

Finding Equivalent Resistance:-

$\boxed{\sf R=R_1+R_2+R_3}$

$\\ \sf\longmapsto R=2+3+5$

$\\ \sf\longmapsto R=10\Omega$

Current:-

According to ohms law.

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{1.5}{10}$

$\\ \sf\longmapsto I=0.15A$