Two resistors of resistances R\(_{1}\) = 100$\Omega$ and R\(_{2}\) = 100$\Omega$ are connected in series. A voltmeter of resistance 400$\Omega$ is connected in parallel to one of the resistance. Find the reading of voltmeter. The emf of battery is 9V. :
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An ideal voltmeter has infinite resistance. If the voltmeter were ideal, the resistance of the parallel branch would remain 100$\Omega$, the total resistance would be 200$\Omega$, the current 9/200 A, and the voltage across R2 would be (9/200)*100 = 4.5V. A real voltmeter has finite resistance and 'loads' the circuit, changing the very voltage it is trying to measure. Its reading will always be lower than the true voltage.
Step 1: Understanding the Question:
We have a simple series circuit with two resistors. A voltmeter, which itself has resistance, is connected across one of the resistors. We need to find the voltage reading shown by this non-ideal voltmeter. Step 2: Key Formula or Approach:
1. Calculate the equivalent resistance of the parallel combination of the resistor and the voltmeter.
2. Calculate the total equivalent resistance of the entire circuit.
3. Use Ohm's law (\(I = V/R_{total}\)) to find the total current flowing from the battery.
4. The voltmeter reading will be the potential difference across the parallel combination, which can be found using \(V_{reading} = I \times R_{parallel}\). Step 3: Detailed Explanation:
Let the voltmeter with resistance \(R_V = 400\Omega\) be connected in parallel with resistor \(R_2 = 100\Omega\). Step 1: Equivalent resistance of the parallel part (\(R_p\)):
\[ \frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_V} = \frac{1}{100} + \frac{1}{400} = \frac{4+1}{400} = \frac{5}{400} = \frac{1}{80} \]
\[ R_p = 80 \Omega \]
Step 2: Total equivalent resistance of the circuit (\(R_{eq}\)):
The resistor \(R_1 = 100\Omega\) is in series with this parallel combination.
\[ R_{eq} = R_1 + R_p = 100\Omega + 80\Omega = 180 \Omega \]
Step 3: Total current (I):
The EMF of the battery is E = 9V.
\[ I = \frac{E}{R_{eq}} = \frac{9 \text{ V}}{180 \Omega} = \frac{1}{20} \text{ A} \]
Step 4: Voltmeter Reading (\(V_{reading}\)):
The voltmeter reads the potential difference across the parallel combination \(R_p\). This current I flows through \(R_p\).
\[ V_{reading} = I \times R_p = \frac{1}{20} \text{ A} \times 80 \Omega = 4 \text{ V} \]
Step 4: Final Answer:
The reading of the voltmeter is 4 V.
