Question
Question: The magnitude of \[i\] ampere units is 
Apply Kirchhoff’s voltage law to the upper loop of the circuit.
−60i+5i1+15i1=0
⇒−60i+20i1=0
⇒20i1=60i
⇒i1=3i
Apply Kirchhoff’s voltage law to the lower loop of the circuit.
−10i2+5i1+15i1=0
−10i2+20i1=0
⇒i2=2i1
Substitute 2i1 for i2 in equation (1).
i+i1+2i1=1A
⇒i+3i1=1A
Substitute 3i for i1 in the above equation.
⇒i+3(3i)=1A
⇒10i=1A
∴i=0.1A
Hence, the current i through the circuit is 0.1A.
Additional information:
According to Kirchhoff current law the sum of all the currents flowing into a junction or node is equal to the sum of all the currents coming out of the junction.According to Kirchhoff’s voltage law the sum of all the voltage drops into a loop is equal to zero.
Note: The students should be careful while applying Kirchhoff’s voltage law to the loops in the circuit. The students should always remember that the sign of the voltage term will be positive if the direction of the loop is opposite to the direction of current and negative if the direction of loop is along the direction of the current.