Four series resistors are connected across a 30 V source and carry 0.125mA . If R1=10$\mathrm { kO }$, R2=33 $\mathrm { k } \Omega$ and R3=47 $\mathrm { k } \Omega$, what is the value of R4 ?
A) 15 $\mathrm { k } \Omega$
B) 150$\Omega$
C) 150$\mathrm { kQ }$
D) 1.5$\mathrm { k } \Omega$

Question

Four series resistors are connected across a  30 V  source and carry  0.125mA .  If  R1=10$\mathrm { kO }$, R2=33  $\mathrm { k } \Omega$ and  R3=47  $\mathrm { k } \Omega$, what is the value of  R4  ?
A) 15 $\mathrm { k } \Omega$
B)  150$\Omega$
C)  150$\mathrm { kQ }$
D)  1.5$\mathrm { k } \Omega$

Quizplus · Accepted Answer

First, convert the current to amperes: 0.125 mA = 0.000125 A. The total resistance $R_{total}$ can be found using Ohm's law: $V = IR$, so $R_{total} = \frac{V}{I} = \frac{30V}{0.000125A} = 240,000 \Omega$ or $240 k\Omega$. Given $R1 = 10 k\Omega$, $R2 = 33 k\Omega$, and $R3 = 47 k\Omega$, the sum of these resistances is $90 k\Omega$. Subtracting this from the total resistance gives $R4 = 240 k\Omega - 90 k\Omega = 150 k\Omega$.

Four Series Resistors Are Connected Across a 30 V kO\mathrm { kO }kO

Four Series Resistors Are Connected Across a 30 V $\mathrm { kO }$