Topic: Networks, Signals and Systems
Resistors R1, R2 and R3 are connected in series across a voltage source 'V' and R4 is connected in parallel with R3. R1 to R4 are to be chosen from available resistor values 3 $\Omega$, 6 $\Omega$, 10 $\Omega$ and 15 $\Omega$. What will be the ratio of maximum current to minimum current through R1 by choosing resistor values appropriately?
- A. 1.8
- B. 3.6
- C. 5.4
- D. 2.7
Answer: A. 1.8
Explanation: The equivalent resistance of the circuit is $R_{eq} = R_1 + R_2 + (R_3 \vert{}\vert{} R_4)$. The total current flows through $R_1$, so $I = V / R_{eq}$.
To maximize current, we must minimize $R_{eq}$. We use the lowest values for the series resistors $R_1, R_2 = 3\Omega, 6\Omega$, and the highest for parallel resistors $R_3, R_4 = 10\Omega, 15\Omega$.
$R_{eq, min} = 3 + 6 + (10 \vert{}\vert{} 15) = 9 + \frac{150}{25} = 9 + 6 = 15 \Omega$.
To minimize current, we must maximize $R_{eq}$. We use the highest values for the series resistors $R_1, R_2 = 10\Omega, 15\Omega$, and the lowest for parallel resistors $R_3, R_4 = 3\Omega, 6\Omega$.
$R_{eq, max} = 10 + 15 + (3 \vert{}\vert{} 6) = 25 + \frac{18}{9} = 25 + 2 = 27 \Omega$.
Ratio of maximum to minimum current = $\frac{I_{max}}{I_{min}} = \frac{V / R_{eq, min}}{V / R_{eq, max}} = \frac{R_{eq, max}}{R_{eq, min}} = \frac{27}{15} = 1.8$.