Question

Two cells of EMF 1.5V & 2.0V & internal resistance of 1& 2 respectively are connected in parallel to an internal resistance of 5 Calculate the current in cach of the three branches of the network ?​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\implies\textsf {Two cells have EMF 1.5V and 2V }\\\implies\textsf{ Internal resistances are \sf 1\Omega \ and \ 2\Omega respectively.}\\\implies\textsf{ They are connected to a resistance of \sf 5\Omega }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The current in each three branches of network. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

For the circuit diagram refer to the attachment.

According to Kirchhoff's Voltage Law , the sum of all voltages around a closed loop in any circuit must be equal to zero , hence here ; The net electromotive force around a closed circuit loop is equal to the sum of potential drops around the loop .

Let $\sf I_1$ be the current through 1.5V cell ( Internal resistance = 1Ω) , and $\sf I_2$ be the current through 2V cell ( Internal resistance= 2Ω) .Then $\sf I_1+I_2$ will be the current through 5Ω resistor.

$\underline{\purple{\boldsymbol{ Using \ KVL \ in \ FEBCF :- }}}$

$\sf:\implies\pink{ 2I_2 + 5(I_1+I_2) - 2 = 0 }\\\\\sf:\implies 2I_2 + 5I_1 + 5I_2 = 2 \\\\\sf:\implies \boxed{\sf 5I_1 + 7I_2 = 2 }$

$\rule{200}2$

$\underline{\purple{\boldsymbol{ Using \ KVL \ in \ ABCDA :- }}}$

$\sf:\implies\pink{ 1(I_1) - 2(I_2) = 1.5 - 2 }\\\\\sf:\implies 1I_1-2I_2 = -0.5 \\\\\sf:\implies \boxed{\sf 2I_2-I_1=0.5 }$

$\rule{200}2$

$\sf:\implies (5I_1+7I_2)+5(2I_2-I_1)=2+5(0.5)\qquad\bigg\lgroup \red{\bf Multipling \ 2nd \ equ^n \ by \ 5} \bigg\rgroup\\\\\sf:\implies 5I_1+7I_2+10I_1-5I_1 = 4.5 \\\\\sf:\implies 17I_2 = 4.5\\\\\sf:\implies I_2=\dfrac{45}{170} \\\\\sf:\implies \boxed{\pink{\mathfrak{ I_2 = \dfrac{ 9}{34}\ Ampere }}}$

$\rule{200}2$

• Finding I1 :-

$\sf:\implies \pink{ 2I_2-I_1=0.5 }\\\\\sf:\implies 2\times\dfrac{9}{34}-I_1=\dfrac{1}{2}\\\\\sf:\implies \dfrac{9}{17}-I_1= \dfrac{1}{2}\\\\\sf:\implies I_1=\dfrac{1}{2}-\dfrac{9}{34}\\\\\sf:\implies I_1=\dfrac{17-9}{34}\\\\\sf:\implies I_1=\dfrac{8}{34} \\\\\sf:\implies \boxed{\pink{\mathfrak{ I_1= \dfrac{ 4}{17}\ Ampere }}}$

$\rule{200}2$

• Hence ,

$\boxed{\bullet \sf{ Current \ through\ DA = I_1 =\pink{\dfrac{4}{17}\: Ampere}}}$

$\boxed{\bullet \sf{ Current \ through\ CB = I_2=\pink{\dfrac{9}{34}\: Ampere}}}$

$\boxed{\bullet \sf{ Current \ through\ FE = (I_1+I_2) =\pink{\dfrac{1}{2}\: Ampere}}}$