Find the voltage drop across each resistor in the circuit

Question

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?

Answer 1

Yes your answer is correct.

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.

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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\$

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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.