In the Guided Problem Solving Problem 3 series circuit with a 9 V source and resistors $R_1 = 3\,k\Omega$ (points 1-2), $R_2 = 10\,k\Omega$ (points 2-3), and $R_3 = 5\,k\Omega$ (points 3-4), what are (a) total resistance, (b) total current, (c) each resistor current, and (d) each resistor voltage?

What the circuit setup tells you

All three resistors ($R_1$, $R_2$, $R_3$) are connected end-to-end from point 1 to point 4, with a $9\, V$ source across points 1 and 4. That is a series circuit, so:

• Resistances add directly.
• The same current flows through every resistor.
• Individual voltage drops add up to the source voltage.

(a) Total resistance for series resistors

For series resistors: $$ R_T = R_1 + R_2 + R_3 $$ Substitute values: $$ R_T = 3\, k\Omega + 10\, k\Omega + 5\, k\Omega = 18\, k\Omega $$

(b) Total current from Ohm's law

Use $I = V/R$ with the source voltage across the total resistance: $$ I_T = \frac{V}{R_T} = \frac{9\, V}{18\,000\,\Omega} = 0.0005\, A = 0.5\, mA $$

(c) Individual currents in a series circuit

In series, there is only one path for current, so: $$ I_1 = I_2 = I_3 = I_T = 0.5\, mA $$

(d) Voltage drops across each resistor

Compute each drop with $V_n = I\, R_n$ (using $I = 0.5\, mA$):

• Across $R_1 = 3\, k\Omega$: $$ V_1 = (0.5\, mA)(3\, k\Omega) = 1.5\, V $$

• Across $R_2 = 10\, k\Omega$: $$ V_2 = (0.5\, mA)(10\, k\Omega) = 5.0\, V $$

• Across $R_3 = 5\, k\Omega$: $$ V_3 = (0.5\, mA)(5\, k\Omega) = 2.5\, V $$

Quick check (Kirchhoff’s Voltage Law for a series loop): $$ V_1 + V_2 + V_3 = 1.5 + 5.0 + 2.5 = 9.0\, V $$ The drops sum to the source voltage, so the results are consistent.