NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
355997
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb
1 increases
2 becomes zero
3 decreases
4 remains same
Explanation:
When dielectric is inserted, \(X_{c}=\dfrac{1}{\omega C}\) decreases, as capacitance increases due to insertion of dielectric and then \(Z=\sqrt{X_{c}^{2}+R^{2}}\) decreases, so current \(\left(I=\dfrac{V}{Z}\right)\) increases and brightness of bulb increases. So, correct option is (1).
JEE - 2024
PHXII07:ALTERNATING CURRENT
355998
The impedance of a \(R - C\) circuit is \({Z_1}\) for a frequency \(f\) and \({Z_2}\) for frequency \(2f\). Then \({Z_1}/{Z_2}\) is
355999
A \({110 {~V}, 60 {~W}}\) lamp is run from a \(220\,V \) ac mains using a capacitor in series with the lamp instead of a resistor, then the voltage across the capacitor is about
355997
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb
1 increases
2 becomes zero
3 decreases
4 remains same
Explanation:
When dielectric is inserted, \(X_{c}=\dfrac{1}{\omega C}\) decreases, as capacitance increases due to insertion of dielectric and then \(Z=\sqrt{X_{c}^{2}+R^{2}}\) decreases, so current \(\left(I=\dfrac{V}{Z}\right)\) increases and brightness of bulb increases. So, correct option is (1).
JEE - 2024
PHXII07:ALTERNATING CURRENT
355998
The impedance of a \(R - C\) circuit is \({Z_1}\) for a frequency \(f\) and \({Z_2}\) for frequency \(2f\). Then \({Z_1}/{Z_2}\) is
355999
A \({110 {~V}, 60 {~W}}\) lamp is run from a \(220\,V \) ac mains using a capacitor in series with the lamp instead of a resistor, then the voltage across the capacitor is about
355997
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb
1 increases
2 becomes zero
3 decreases
4 remains same
Explanation:
When dielectric is inserted, \(X_{c}=\dfrac{1}{\omega C}\) decreases, as capacitance increases due to insertion of dielectric and then \(Z=\sqrt{X_{c}^{2}+R^{2}}\) decreases, so current \(\left(I=\dfrac{V}{Z}\right)\) increases and brightness of bulb increases. So, correct option is (1).
JEE - 2024
PHXII07:ALTERNATING CURRENT
355998
The impedance of a \(R - C\) circuit is \({Z_1}\) for a frequency \(f\) and \({Z_2}\) for frequency \(2f\). Then \({Z_1}/{Z_2}\) is
355999
A \({110 {~V}, 60 {~W}}\) lamp is run from a \(220\,V \) ac mains using a capacitor in series with the lamp instead of a resistor, then the voltage across the capacitor is about
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII07:ALTERNATING CURRENT
355997
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb
1 increases
2 becomes zero
3 decreases
4 remains same
Explanation:
When dielectric is inserted, \(X_{c}=\dfrac{1}{\omega C}\) decreases, as capacitance increases due to insertion of dielectric and then \(Z=\sqrt{X_{c}^{2}+R^{2}}\) decreases, so current \(\left(I=\dfrac{V}{Z}\right)\) increases and brightness of bulb increases. So, correct option is (1).
JEE - 2024
PHXII07:ALTERNATING CURRENT
355998
The impedance of a \(R - C\) circuit is \({Z_1}\) for a frequency \(f\) and \({Z_2}\) for frequency \(2f\). Then \({Z_1}/{Z_2}\) is
355999
A \({110 {~V}, 60 {~W}}\) lamp is run from a \(220\,V \) ac mains using a capacitor in series with the lamp instead of a resistor, then the voltage across the capacitor is about
