In the given diagram #E_1# and #E_2# are the emfs of cells, #r_1# and#r_2# are the internal resistances of cells. #R# is the external resistance. What are the values of #I_1# and #I_2#?

Question

Nora Arzate · Accepted Answer

Applying Kirchhoff's voltage law in the loop containing #E_1,r_1 and R# we get

#R(I_1+I_2)+r_1I_1-E_1=0#

#=>12(I_1+I_2)+0.5*I_1-6=0#

#=>24I_1+24I_2+I_1-12=0#

#=>25I_1+24I_2=12.... [1]#

Applying Kirchhoff's voltage law in the loop containing #E_2,r_2 and R# we get

#R(I_1+I_2)+r_2I_1-E_2=0#

#=>12(I_1+I_2)+1*I_2-10=0#

#=>12I_1+12I_2+I_2-10=0#

#=>12I_1+13I_2=10.... [2]#

Multiplying [1] by 12 and [2] by 25 we get

#300I_1+288I_2=144...[3]#

#300I_1+325I_2=250.... [4]#

Subtracting [3] from [4] we get

#37I_2=106#

#=>I_2=106/37A~~2.86A#

Inserting #I_2=106/37# in [2]

#=>12I_1+13*106/37=10#

#I_1=(370-13*106)/(37*12)A~~-2.27A#

Isabella Alvarez · Answer

undefined Using Kirchhoff's laws, we can analyze the circuit. Applying Kirchhoff's voltage law to the outer loop:

$$E_1 - I_1(r_1 + R) - E_2 = 0$$

Similarly, for the inner loop:

$$E_2 - I_2(r_2 + R) = 0$$

Solving these two equations will give us the values of $I_1$ and $I_2$. We can rearrange the equations as needed and then solve them simultaneously.