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 \( 12 \Omega \). \[ (Z=13 \Omega) \]

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!