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Difficulty: MediumSeries Circuits

A railway signal monitoring circuit is powered by a 72 V72\text{ V} DC power supply. The circuit consists of three components connected in series: an 85 Ω85\ \Omega relay coil, a 35 Ω35\ \Omega surge-limiting resistor, and an unknown protective resistor RxR_x. If the entire circuit draws a steady current of 0.4 A0.4\text{ A}, what is the resistance of the protective resistor RxR_x in ohms (Ω\Omega)?

Answer: 60 Ω

Answer

The resistance of the protective resistor RxR_x is 60 Ω60\ \Omega.
Applying Ohm's Law (RT=VTIR_T = \frac{V_T}{I}) gives a total series resistance of 180 Ω180\ \Omega. Since resistances in series add directly (RT=85 Ω+35 Ω+Rx=120 Ω+RxR_T = 85\ \Omega + 35\ \Omega + R_x = 120\ \Omega + R_x), solving for the unknown yields Rx=180 Ω−120 Ω=60 ΩR_x = 180\ \Omega - 120\ \Omega = 60\ \Omega.

Step-by-Step Solution

1
Calculate the total circuit resistance (RTR_T) using Ohm's Law.
RT=VsourceItotal=72 V0.4 A=180 ΩR_T = \frac{V_{\text{source}}}{I_{\text{total}}} = \frac{72\text{ V}}{0.4\text{ A}} = 180\ \Omega
In a series circuit, total resistance equals total source voltage divided by total circuit current.
2
Determine the combined resistance of the known series components.
Rknown=Rrelay+Rsurge=85 Ω+35 Ω=120 ΩR_{\text{known}} = R_{\text{relay}} + R_{\text{surge}} = 85\ \Omega + 35\ \Omega = 120\ \Omega
Resistances in a series circuit are directly additive.
3
Subtract the known series resistance from the total circuit resistance to solve for RxR_x.
Rx=RT−Rknown=180 Ω−120 Ω=60 ΩR_x = R_T - R_{\text{known}} = 180\ \Omega - 120\ \Omega = 60\ \Omega
The sum of all individual series resistors must equal the total equivalent resistance (RT=R1+R2+RxR_T = R_1 + R_2 + R_x).

Key Concept

In a direct-current (DC) series circuit, current is identical through every component (IT=I1=I2=I3I_T = I_1 = I_2 = I_3), and the total equivalent resistance is the arithmetic sum of all individual resistances (RT=R1+R2+⋯+RnR_T = R_1 + R_2 + \dots + R_n).

Alternative Method

Find the voltage drops across the known components first: Vrelay=0.4 A×85 Ω=34 VV_{\text{relay}} = 0.4\text{ A} \times 85\ \Omega = 34\text{ V} and Vsurge=0.4 A×35 Ω=14 VV_{\text{surge}} = 0.4\text{ A} \times 35\ \Omega = 14\text{ V}. Subtract these from the total source voltage using Kirchhoff's Voltage Law: VRx=72 V−(34 V+14 V)=24 VV_{Rx} = 72\text{ V} - (34\text{ V} + 14\text{ V}) = 24\text{ V}. Then find Rx=VRxI=24 V0.4 A=60 ΩR_x = \frac{V_{Rx}}{I} = \frac{24\text{ V}}{0.4\text{ A}} = 60\ \Omega.
Estimated Time:1m 0s
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Series Circuits

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