5.2 Find the current which would flow in each of the above cases if the applied voltage was 10 V . $$ \begin{array}{l} \text { (at } 100 \mathrm{~Hz}, \mathrm{I}=6,3 \mathrm{ma} \ \text { at } 5000 \mathrm{~Hz}, \mathrm{I}=314 \mathrm{~m} \end{array} $$ $$ (\text { at } 5000 \mathrm{~Hz}, I=6,3 \mathrm{~mm} \text { ) } $$ 6. What value of capacitance will have a reactance of $$ 3180 \Omega $$ when connected to a 600 the supply? 7. In a series RL circuit, under what circumstances would $$ \phi $$ be $$ 0^{\circ} $$ or $$ 90^{\circ} $$ ? ( $$ \phi $$ will be zero if the circuit is purely resistive) ( $$ \phi $$ will be $$ 90^{\circ} $$ if the circuit is purely inductive). 8. In a series $$ R C $$ circuit, under what circumstances would $$ \phi $$ be exactly $$ 0^{\circ} $$ or $$ -90^{\circ} $$ ? ( $$ \phi $$ will be zero if the circuit is purely resistive) ( $$ \phi $$ will be $$ -90^{\circ} $$ if the circuit is purely capacitive). 9. Calculate the inductance of an inductor which has a reactance of $$ 377 \Omega $$ when connected to a 120 Hz supply. $$ (L=0,5 H) $$ 10. At what frequency will a coil of inductance 80 mH have a reactance of $$ 302 \Omega $$ ? $$ (f=600 \mathrm{~Hz}) $$ 11. Calculate the capacitance of a capacitor which has a reactance of $$ 36 \Omega $$ when connected to a 50 Hz supply. $$ (C=88,4 \mu F) $$ 12. At what frequency will a capacitor of 1000 pF have a reactance of $$ 19890 \Omega $$ ? ( $$ f=8 \mathrm{kHz} $$ ) 13. Find the impedance of a coil which has a resistance of $$ 5 \Omega $$ and a reactance of $$ 12 \Omega $$. $$ (Z=13 \Omega) $$

Question

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Let's solve the problems step by step.

### Problem 5: Current Calculation at Different Frequencies

We need to find the current $$ I $$ when the applied voltage $$ V = 10 \, V $$ for two cases:

1. At $$ 100 \, Hz $$ where $$ I = 6.3 \, mA $$
2. At $$ 5000 \, Hz $$ where $$ I = 314 \, mA $$

Using Ohm's Law, $$ I = \frac{V}{R} $$, we can find the resistance $$ R $$ for each case and then calculate the new current for $$ V = 10 \, V $$.

#### Case 1: At $$ 100 \, Hz $$

Given:
- $$ I_1 = 6.3 \, mA = 0.0063 \, A $$

Using Ohm's Law:
$$
R_1 = \frac{V}{I_1} = \frac{10}{0.0063}
$$

#### Case 2: At $$ 5000 \, Hz $$

Given:
- $$ I_2 = 314 \, mA = 0.314 \, A $$

Using Ohm's Law:
$$
R_2 = \frac{V}{I_2} = \frac{10}{0.314}
$$

Now, we can calculate the resistances and then find the new currents for $$ V = 10 \, V $$.

### Problem 6: Capacitance Calculation

We need to find the capacitance $$ C $$ that will have a reactance $$ X_C = 3180 \, \Omega $$ when connected to a $$ 600 \, Hz $$ supply.

The formula for capacitive reactance is:
$$
X_C = \frac{1}{2 \pi f C}
$$

Rearranging for $$ C $$:
$$
C = \frac{1}{2 \pi f X_C}
$$

Substituting $$ f = 600 \, Hz $$ and $$ X_C = 3180 \, \Omega $$.

### Problem 9: Inductance Calculation

We need to find the inductance $$ L $$ of an inductor which has a reactance $$ X_L = 377 \, \Omega $$ when connected to a $$ 120 \, Hz $$ supply.

The formula for inductive reactance is:
$$
X_L = 2 \pi f L
$$

Rearranging for $$ L $$:
$$
L = \frac{X_L}{2 \pi f}
$$

Substituting $$ X_L = 377 \, \Omega $$ and $$ f = 120 \, Hz $$.

### Problem 11: Capacitance Calculation

We need to find the capacitance $$ C $$ of a capacitor which has a reactance $$ X_C = 36 \, \Omega $$ when connected to a $$ 50 \, Hz $$ supply.

Using the same formula as in Problem 6:
$$
C = \frac{1}{2 \pi f X_C}
$$

Substituting $$ f = 50 \, Hz $$ and $$ X_C = 36 \, \Omega $$.

### Problem 13: Impedance Calculation

We need to find the impedance $$ Z $$ of a coil which has a resistance $$ R = 5 \, \Omega $$ and a reactance $$ X = 12 \, \Omega $$.

The formula for impedance is:
$$
Z = \sqrt{R^2 + X^2}
$$

Now, let's perform the calculations for all the problems.

### Performing Calculations

I'll calculate the required values now.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{377}{\left(2\pi \times 120\right)}$$
- step1: Remove the parentheses:
  $$\frac{377}{2\pi \times 120}$$
- step2: Multiply the terms:
  $$\frac{377}{240\pi }$$
Calculate or simplify the expression $$ \sqrt(5^2 + 12^2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{5^{2}+12^{2}}$$
- step1: Add the numbers:
  $$\sqrt{169}$$
- step2: Write in exponential form:
  $$\sqrt{13^{2}}$$
- step3: Simplify the root:
  $$13$$
Calculate or simplify the expression $$ 10/0.314 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{10}{0.314}$$
- step1: Convert the expressions:
  $$\frac{10}{\frac{157}{500}}$$
- step2: Multiply by the reciprocal:
  $$10\times \frac{500}{157}$$
- step3: Multiply:
  $$\frac{10\times 500}{157}$$
- step4: Multiply:
  $$\frac{5000}{157}$$
Calculate or simplify the expression $$ 10/0.0063 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{10}{0.0063}$$
- step1: Convert the expressions:
  $$\frac{10}{\frac{63}{10000}}$$
- step2: Multiply by the reciprocal:
  $$10\times \frac{10000}{63}$$
- step3: Multiply:
  $$\frac{10\times 10000}{63}$$
- step4: Multiply:
  $$\frac{100000}{63}$$
Calculate or simplify the expression $$ 1/(2*\pi*600*3180) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1}{\left(2\pi \times 600\times 3180\right)}$$
- step1: Remove the parentheses:
  $$\frac{1}{2\pi \times 600\times 3180}$$
- step2: Multiply the terms:
  $$\frac{1}{3816000\pi }$$
Calculate or simplify the expression $$ 1/(2*\pi*50*36) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1}{\left(2\pi \times 50\times 36\right)}$$
- step1: Remove the parentheses:
  $$\frac{1}{2\pi \times 50\times 36}$$
- step2: Multiply the terms:
  $$\frac{1}{3600\pi }$$
Here are the detailed calculations and results for each problem:

### Problem 5: Current Calculation at Different Frequencies

1. **At $$ 100 \, Hz $$**:
   $$
   R_1 = \frac{10}{0.0063} \approx 1587.3 \, \Omega
   $$
   Now, using this resistance to find the current at $$ 10 \, V $$:
   $$
   I_1 = \frac{10}{1587.3} \approx 0.0063 \, A \text{ (remains the same)}
   $$

2. **At $$ 5000 \, Hz $$**:
   $$
   R_2 = \frac{10}{0.314} \approx 31.85 \, \Omega
   $$
   Now, using this resistance to find the current at $$ 10 \, V $$:
   $$
   I_2 = \frac{10}{31.85} \approx 0.313 \, A
   $$

### Problem 6: Capacitance Calculation

To find the capacitance $$ C $$ that will have a reactance of $$ 3180 \, \Omega $$ at $$ 600 \, Hz $$:
$$
C = \frac{1}{2 \pi (600)(3180)} \approx 8.34 \times 10^{-8} \, F \text{ or } 83.4 \, nF
$$

### Problem 9: Inductance Calculation

To find the inductance $$ L $$ with a reactance of $$ 377 \, \Omega $$ at $$ 120 \, Hz $$:
$$
L = \frac{377}{2 \pi (120)} \approx 0.5 \, H
$$

### Problem 11: Capacitance Calculation

To find the capacitance $$ C $$ with a reactance of $$ 36 \, \Omega $$ at $$ 50 \, Hz $$:
$$
C = \frac{1}{2 \pi (50)(36)} \approx 8.84 \times 10^{-5} \, F \text{ or } 88.4 \, \mu F
$$

### Problem 13: Impedance Calculation

To find the impedance $$ Z $$ of a coil with resistance $$ 5 \, \Omega $$ and reactance $$ 12 \, \Omega $$:
$$
Z = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \Omega
$$

### Summary of Results
- **Current at $$ 100 \, Hz $$**: $$ 6.3 \, mA $$
- **Current at $$ 5000 \, Hz $$**: $$ 313 \, mA $$
- **Capacitance for $$ 3180 \, \Omega $$**: $$ 83.4 \, nF $$
- **Inductance for $$ 377 \, \Omega $$**: $$ 0.5 \, H $$
- **Capacitance for $$ 36 \, \Omega $$**: $$ 88.4 \, \mu F $$
- **Impedance**: $$ 13 \, \Omega $$

If you have any further questions or need additional calculations, feel free to ask!
- **Current at 100 Hz**: 6.3 mA - **Current at 5000 Hz**: 313 mA - **Capacitance for 3180 Ω**: 83.4 nF - **Inductance for 377 Ω**: 0.5 H - **Capacitance for 36 Ω**: 88.4 μF - **Impedance**: 13 Ω

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