Given below are two statements : One is labelled as Assertion \(A\) and the other is labelled as Reason R
Assertion A : For measuring the potential difference across a resistance of \(600 \Omega\), the voltmeter with resistance \(1000 \Omega\) will be preferred over voltmeter with resistance \(4000 \Omega\).
Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below
Remember that an ideal voltmeter has infinite resistance. A practical voltmeter should have as high a resistance as possible to minimize the impact on the circuit being measured.
Both \(A\) and \(R\) are correct but \(R\) is not the correct explanation of \(A\)
Both \(A\) and \(R\) are correct and \(R\) is the correct explanation of \(A\)
A is not correct but \(R\) is correct
A is correct but \(R\) is not correct
Step 1: Analyze Assertion A
An ideal voltmeter has infinite resistance. A real voltmeter should have a resistance much higher than the resistance it is measuring across to minimize the current drawn by the voltmeter and ensure an accurate reading of the potential difference. In this case, measuring across a 600 Ω resistor, a voltmeter with 4000 Ω resistance is preferred over a voltmeter with 1000 Ω resistance because it’s closer to the ideal. Thus, the assertion is incorrect.
Step 2: Analyze Reason R
A voltmeter is connected in parallel to the resistor it is measuring. According to Ohm's law, \( V = IR \), the current (\( I \)) drawn by a voltmeter is inversely proportional to its resistance (\( R \)) if the voltage (\( V \)) is constant. Therefore, a voltmeter with higher resistance will draw a smaller current. Thus, the reason is correct.
Conclusion: Assertion A is incorrect, but Reason R is correct. Thus, the correct option is (3).
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.
Match List-I with List-II: List-I
Current electricity is defined as the flow of electrons from one section of the circuit to another.
There are two types of current electricity as follows:
The current electricity whose direction remains the same is known as direct current. Direct current is defined by the constant flow of electrons from a region of high electron density to a region of low electron density. DC is used in many household appliances and applications that involve a battery.
The current electricity that is bidirectional and keeps changing the direction of the charge flow is known as alternating current. The bi-directionality is caused by a sinusoidally varying current and voltage that reverses directions, creating a periodic back-and-forth motion for the current. The electrical outlets at our homes and industries are supplied with alternating current.