Find the voltage drop across each resistor in the circuit

I have this question to do.

The current through LED is $7.5\ \rm mA$. Find voltage drop across each resistor in the circuit and value of $R_3$. Verify KVL in the circuit I have personally solved the question. Here is my solution:

I have current and current is always same in series circuit. I used that current value to find to find voltage drop across each resistor. When I found that I used Kirchhoff's law $$V_s = V_{R_1} + V_{R_2} + V_{R_3} + V_d$$ to find $V_{R_3}$ and then using that I found $R_3$.

\begin{align} V_{R_1}&=8.25\ \rm V\\ V_{R_2}&=3.375\ \rm V\\ V_d&=2.1\ \rm V\\ V_s&=V_{R_1}+V_{R_2}+V_d+V_{R_3}\\ &20-8.25-3.375-2.1=V_{R_3}\\ V_{R_3}&=6.275\ \rm V\\ V_{R_3}&=IR_3\\ &\frac{6.275}{7.5\times10^{-3}}=R_3\\ R_3&=836.67\ \rm\Omega \end{align}

Is my solution correct?

• Looks good R3 = (20V - 2.1V)/7.5mA - 1.1kΩ - 450Ω = 836.67Ω = 820Ω
– G36
Mar 18 '18 at 14:49
• just a snide side remark: I don't like educational material that mistypes units. It's "kW" (1000 Watt), not "KW" (Kelvin·Watt) Mar 18 '18 at 15:15
• @MarcusMüller did you comment on the wrong question? I see no KW? Mar 18 '18 at 15:19
• Sorry, meant KΩ, @Trevor_G Mar 18 '18 at 15:22
• @MarcusMüller ah... That's a common one for sure. Mar 18 '18 at 15:23

I would however have shown the equations for finding the first two voltages. In a test the method usually gets you more marks than the value.

BTW: There are almost always more than one way to solve these kinds of problems. As such, it is prudent, and a good habit to get into, to do so to verify your work.

In this example. you know

$R_{TOTAL} = (V_S-V_d)/I_D = (20-2.1)/0.0075 = 2386.67\Omega$

So

$R_3 = 2386.67 - 1100 - 450 = 836.67\Omega$

Addition: Since the assignment also asks you to verify KVL, it may be prudent to work your answer the other way around. Using KVL as part of your initial math method does not "prove" it per se.

So finding R3 from the above, then calculating the voltages across reach resistor to show that $V_{R1} + V_{R2} + V_{R3} + V_D = V_S$ would be a better way to answer it.

• Trevor the genius! ;^) Mar 18 '18 at 15:09
• Blushing...... Always nice to hear though :D Mar 18 '18 at 15:12
• @AndrewMorton hmm.. thanks. I wonder how many folks know to read it as Latin that way though... Mar 19 '18 at 13:58