An oscilloscope has an input resistance of \( 1\, {M}\Omega \). A 10X passive attenuating probe is connected to it to increase the input voltage range as well as the effective input resistance. The effective input resistance, in \( {M}\Omega \), seen into the probe tip is:
Show Hint
- \( 0.9 \)
- \( 9.1 \)
- \( 10 \)
- \( 11 \)
The Correct Option is C
Solution and Explanation
Step 1: Understanding the probe configuration.
A 10X passive probe attenuates the voltage by a factor of 10. This is achieved using a resistor in the probe (say \( R_p \)) and the oscilloscope's internal input resistance \( R_{{scope}} \) arranged in a voltage divider.
Step 2: Determine the resistor values.
To get 10X attenuation: \[ \frac{R_{{scope}}}{R_p + R_{{scope}}} = \frac{1}{10} \Rightarrow \frac{1\,{M}\Omega}{R_p + 1\,{M}\Omega} = \frac{1}{10} \] Solving: \[ 10 = \frac{R_p + 1}{1} \Rightarrow R_p = 9\,{M}\Omega \] Step 3: Effective input resistance seen into the probe tip.
The total resistance from the probe tip into the oscilloscope is the series combination of: \[ R_{{effective}} = R_p + R_{{scope}} = 9 + 1 = 10\,{M}\Omega \]
Top Questions on Measurements
- A metallic strain-gauge (SG) with resistance \( R_{SG} \) is connected as shown in the figure, where \( R_{L1} \), \( R_{L2} \), \( R_{L3} \) represent the lead wire resistances. The SG has a gauge factor of 2 and nominal resistance \( R_N \) of \( 125 \, \Omega \). When the SG is subjected to a tensile strain of \( 2 \times 10^{-3} \), the resulting change in \( R_{SG} \) is \( \Delta R \). The \( \Delta R \) value is measured as \( \Delta R_{MEAS} = R_{EQ2} - R_{EQ1} \). The \( R_{EQ1} \) and \( R_{EQ2} \) are the equivalent resistances measured between the terminals 1 and 2, and terminals 2 and 3, respectively.
If \( R_{L1} = R_{L2} = 5 \, \Omega \), and \( R_{L3} = 4.95 \, \Omega \), the measured value of tensile strain is ________ \( \times 10^{-3} \) (rounded off to two decimal places). - Power consumed by a three-phase balanced load is measured using two-wattmeter method. The per-phase average power drawn by the load is 30 kW at \( \frac{\sqrt{3}}{2} \) lagging power factor. The readings of the wattmeters will be
Match the following sensors with their most suitable applications.
In a single-phase AC circuit, the power consumed by load resistance $R_L$ for an excitation $V_S$ is measured by a wattmeter. The same wattmeter is connected in two different topologies, Topology-A and Topology-B, as shown in the figure. Different branch currents and voltage drops are also marked in the figure. Among the following options, the condition that ensures low error in the wattmeter reading for both the topologies is
Questions Asked in GATE IN exam
- If \( P e^x = Q e^{-x} \) for all real values of \( x \), which one of the following statements is true?
- Let the difference equation \( y[n] = \alpha y[n-1] + x[n] \), where \( \alpha>1 \) and \( \alpha \) is real, represent a causal discrete-time linear time invariant system. The system is initially at rest. If \( x[n] = -\delta[n-p] \), where \( p>10 \), the value of \( y[p+1] \) is:
- GATE IN - 2025
- Linear Algebra
- Let a continuous-time signal be \( x(t) = e^{j9t} + e^{j5t} \), where \( j = \sqrt{-1} \) and \( t \) is in seconds. The fundamental period of the magnitude of \( x(t) \), in seconds, is:
- GATE IN - 2025
- Signals
Bird : Nest :: Bee : __________
Select the correct option to complete the analogy.For the circuit shown in the figure, the active power supplied by the source is ________ W (rounded off to one decimal place).
- GATE IN - 2025
- Electrical Circuits