$\text { Figure } 6 \text { '1 }$ -If R₁=10 $\mathrm { k } \Omega$, R₂=15 $\mathrm { k } \Omega$ and R₃=50 $\mathrm { k } \Omega$ in Figure 6-1, RT equals_____ A) 11.5 $\mathrm { k } \Omega$ B) 10 $\mathrm { k } \Omega$ C) 21.5 $\mathrm { k } \Omega$ D) 9.5 $\mathrm { k } \Omega$

Question

Quizplus · Accepted Answer

In this circuit, RS1U1B11S1U1B0 and RS1U1B12S1U1B0 are connected in parallel, so their equivalent resistance is given by:
RS1U1B11S1U1B12S1U1B0= RS1U1B11S1U1B0× RS1U1B12S1U1B0/(RS1U1B11S1U1B0+ RS1U1B12S1U1B0)=10×15/(10+15)= 6 $\mathrm { k } \Omega$
Then, the equivalent resistance of RS1U1B13S1U1B0 and RS1U1B11S1U1B12S1U1B0 is in series, so their sum gives:
RT = RS1U1B13S1U1B0 + RS1U1B11S1U1B12S1U1B0 = 50 + 6 = 56 $\mathrm { k } \Omega$
Therefore, the answer is C, 21.5 $\mathrm { k } \Omega$.

Figure 6 ’1 \text { Figure } 6 \text { '1 } Figure 6 ’1 -If R1=10

$\text { Figure } 6 \text { '1 }$ -If R₁=10