Question

عند توصيل الفولتميتر عبر بطارية قوتها الدافعة الكهربية 30 ومهملة المقاومة الداخلية كما موضح بالشكل الحرف مؤشرة إلى نهاية تدريجه وعند توصيل مقاومة (R) مع مضاعف الجهد داخل الفواتيميتر بينما يتصل بطرفي البطارية الحرف مؤشرة إلى منتصف تدريجه

ما قيمة المقاومة (R) وطريقة توصيلها ؟

• A. 3000 $\Omega$ ، على التوازي.
• B. 3000 $\Omega$ ، على التوالي.
• C. 1500 $\Omega$ ، على التوالي.
• D. 1500 $\Omega$ ، على التوازي.

Answer 1

Answer: (B)

3000 $\Omega$ ، على التوالي.

Answer 2

The problem describes a voltmeter constructed from a galvanometer and a series multiplier resistance. We need to determine the value and connection method of an additional resistance (R) added to the multiplier part, given the change in deflection.

Step 1: Calculate the full-scale deflection current ($I_G$) of the galvanometer.

Initially, the voltmeter consists of a galvanometer with resistance $R_G = 50 \, \Omega$ connected in series with a multiplier resistance $R_M = 2950 \, \Omega$. The total resistance of the voltmeter ($R_{V}$) is: $R_{V} = R_G + R_M = 50 \, \Omega + 2950 \, \Omega = 3000 \, \Omega$

When this voltmeter is connected across a battery with EMF $V = 3 \, V$, the pointer deflects to the end of its scale. This means $3 \, V$ is the full-scale voltage ($V_{max}$) and the current flowing through the galvanometer is the full-scale deflection current ($I_G$). Using Ohm's Law: $V_{max} = I_G \times R_{V}$ $3 \, V = I_G \times 3000 \, \Omega$ $I_G = \frac{3 \, V}{3000 \, \Omega} = 0.001 \, A = 1 \, mA$

Step 2: Determine the new total resistance of the voltmeter when R is added.

When the resistance (R) is added, the voltmeter is still connected to the $3 \, V$ battery, but the pointer deflects to the middle of its scale. This means the current flowing through the galvanometer is half of the full-scale deflection current: $I' = \frac{I_G}{2} = \frac{1 \, mA}{2} = 0.5 \, mA = 0.0005 \, A$

Let the new total resistance of the voltmeter be $R'_{V}$. Using Ohm's Law for this new configuration: $V = I' \times R'_{V}$ $3 \, V = 0.0005 \, A \times R'_{V}$ $R'_{V} = \frac{3 \, V}{0.0005 \, A} = 6000 \, \Omega$

Step 3: Determine the value and connection of R.

The galvanometer resistance $R_G = 50 \, \Omega$ remains unchanged and is always in series with the effective multiplier resistance. The new total resistance of the voltmeter is $R'_{V} = R_G + R'_{M,eff}$, where $R'_{M,eff}$ is the effective resistance of the multiplier part after adding R. $6000 \, \Omega = 50 \, \Omega + R'_{M,eff}$ $R'_{M,eff} = 6000 \, \Omega - 50 \, \Omega = 5950 \, \Omega$

The original multiplier resistance was $R_M = 2950 \, \Omega$. We need to connect R with $R_M$ to get an effective resistance of $R'_{M,eff} = 5950 \, \Omega$.

• If R is connected in parallel with $R_M$: The equivalent resistance of two resistors in parallel is always less than the smallest individual resistance. Since $R_M = 2950 \, \Omega$, a parallel combination with R would result in $R'_{M,eff} < 2950 \, \Omega$. However, we calculated $R'_{M,eff} = 5950 \, \Omega$. Therefore, R cannot be connected in parallel with $R_M$.

• If R is connected in series with $R_M$: The equivalent resistance of two resistors in series is their sum. $R'_{M,eff} = R_M + R$ $5950 \, \Omega = 2950 \, \Omega + R$ $R = 5950 \, \Omega - 2950 \, \Omega = 3000 \, \Omega$

This result is consistent with the requirement. Thus, the resistance R is $3000 \, \Omega$ and it is connected in series with the multiplier resistance.