Question

20 ohm 30 ohm and 50 ohm resistors are conected in series draw the circuit​

Answer 1

Answer:-

Circuit diagram:-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(1,5)\qbezier(1,1)(1,1)(3,1)\qbezier(3,0.5)(3,0.5)(3,1.5)\qbezier(3.2,0.7)(3.2,0.7)(3.2,1.3)\qbezier(3.4,0.5)(3.4,0.5)(3.4,1.5)\qbezier(3.6,0.7)(3.6,0.7)(3.6,1.3)\qbezier(3.8,0.5)(3.8,0.5)(3.8,1.5)\qbezier(4,0.7)(4,0.7)(4,1.3)\qbezier(4,1)(4,1)(11,1)\qbezier(11,1)(11,1)(11,5.1)\qbezier(1,5)(1,5)(1.6,5)\qbezier(1.6,5)(1.6,5)(1.8,5.4)\qbezier(1.8,5.4)(1.8,5.4)(2,4.8)\qbezier(2,4.8)(2,4.8)(2.2,5.4)\qbezier(2.2,5.4)(2.2,5.4)(2.4,4.8)\qbezier(2.4,4.8)(2.4,4.8)(2.6,5.4)\qbezier(2.6,5.4)(2.6,5.4)(2.8,4.8)\qbezier(2.8,4.8)(2.8,4.8)(2.9,5.1)\qbezier(2.9,5.1)(2.9,5.1)(3.5,5.1)\qbezier(3.5,5.1)(3.5,5.1)(3.7,5.4)\qbezier(3.7,5.4)(3.7,5.4)(3.9,4.8)\qbezier(3.9,4.8)(3.9,4.8)(4.1,5.4)\qbezier(4.1,5.4)(4.1,5.4)(4.3,4.8)\qbezier(4.3,4.8)(4.3,4.8)(4.5,5.4)\qbezier(4.5,5.4)(4.5,5.4)(4.7,4.8)\qbezier(4.7,4.8)(4.7,4.8)(4.8,5.1)\qbezier(4.8,5.1)(4.8,5.1)(5.5,5.1)\qbezier(5.5,5.1)(5.5,5.1)(5.7,5.4)\qbezier(5.7,5.4)(5.7,5.4)(5.9,4.8)\qbezier(5.9,4.8)(5.9,4.8)(6.1,5.4)\qbezier(6.1,5.4)(6.1,5.4)(6.3,4.8)\qbezier(6.3,4.8)(6.3,4.8)(6.5,5.4)\qbezier(6.5,5.4)(6.5,5.4)(6.7,4.8)\qbezier(6.7,4.8)(6.7,4.8)(6.8,5.1)\qbezier(6.8,5.1)(6.8,5.1)(11,5.1)\put(3,1){\vector(-1,0){1}}\put(1.9,5.8){ \sf 20\Omega }\put(3.9,5.8){ \sf 30\Omega }\put(5.9,5.8){ \sf 50\Omega }\put(2.1,0.5){\sf I}\put(3.4,-0.3){\sf V}\end{picture}$

Finding Equivalent Resistance:-

$\boxed{\sf R_{(Equivalent)}=R_1+R_2++R_3}$

$\\ \sf\hookrightarrow R_{(Equivalent)}=20+30+50$

$\\ \sf\hookrightarrow R_{(Equivalent)}=100\Omega(ohm)$