ELECTRONICS: Kirchhoff’s Laws

Q: What are the values of the currents ( I ) and unknown voltage drops ( V ) across the resistors ( R ) pictured below?

A: The first problem-solving step involves assigning labels to the junctions ( j ) in the circuit:

We must now sketch the currents flowing in the circuit:

The current directions have been assigned arbitrarily. Positive final current derivations indicate that current directions have been assigned correctly, and vice versa. The loop directions assigned to Loop A and Loop B have been chosen arbitrarily as well:

The junction rule states that the sum of currents ( I ) entering a junction node ( j ) equals the sum of the currents emerging from the same node:

I₀ = I₁ + I₂ 

It is equally important to note that the sum of all voltage ( V ) rises and drops around a closed path ( loop ) is zero. This is consistent with the First Law of Thermodynamics, which states that “ Energy cannot be created or destroyed; it only changes form. “ The high-energy state that the voltage source provides its electrons is completely “ spent “ via work on resistors and heat loss:

            Loop A: V₁ – V₂ = 0                    Loop B: V₁ – V₂ – V₃ = 0

            1.5V – ( I₂ )( 100Ω ) = 0            ( I₂ )( 100Ω ) – 9V – ( I₁ )( 200Ω ) = 0

            I₂ = 0.015A                               ( I₂ )( 100Ω ) – ( I₁ )( 200Ω ) = 9V

We may now substitute the ( I₂ ) value derived from Loop A into the Loop B equation:

( 0.015A )( 100Ω ) – ( I₁ )( 200Ω ) = 9V

-( I₁ )( 200Ω ) = 7.5V

I₁ = -0.0375A

The negative sign tells us that the current direction originally assigned to ( I₁ ) is incorrect; thus, the negative value derived reflects the physics of the circuit, and it is used in our calculation of ( I₀ ):

I₀ = I₁ + I₂ 

I₀ = -0.0375A + 0.015A

I₀ = -0.0225A

The negative value of ( I₀ ) shows that it was assigned an incorrect direction as well. We now have what is needed to solve for the various unknown voltage values around the circuit. The positive value of current is used to determine the value of voltage drops across the circuit ( high to low energy state ):

V_100Ω = ( 0.015A )( 100Ω )

V_100Ω = 1.5V

V_200Ω = ( 0.0375A )( 200Ω )

V_200Ω = 7.5V

Finally, the power dissipated at each resistor is as follows:

P = IV

P_100Ω = ( 0.015A )( 1.5V )

P_100Ω = 0.0225W = 22.5mW

P_200Ω = ( 0.0375A )( 7.5V )

P_200Ω = 0.28125W = 281mW