A piezoelectric transducer having capacitance of 1nF and leakage resistance of 1012 - is connected to a parallel load ZL- ZL comprises of 1 M- resistance RL and 0-5 nF capacitance C1 in parallel- What will be the approximate value of load voltage VL at very high frequency for 1nC of charge-

The load voltage is given as nbsp; On Analysing we find, V_L= R_L jω C_P nbsp;V_o1+jω R_L.(C_P+ C_L) At very high frequency ω nbsp;R_L (C_P + C_L) ...

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Standard XII

Physics

Increasing Capacitance Even More

A piezoelectr...

Question

A piezoelectric transducer having capacitance of 1nF and leakage resistance of $10^{12} Ω$ is connected to a parallel load $Z_{L} .$ $Z_{L}$ comprises of $1 M Ω$ resistance $(R_{L})$ and 0.5 nF capacitance $(C_{1})$ in parallel. What will be the approximate value of load voltage $V_{L}$ at very high frequency for 1nC of charge?

0.667 μ V

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0.667 m V

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0.667 V

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6.67 V

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Solution

The correct option is C $0.667 V$
The load voltage is given as

On Analysing we find,

$V_{L} = \frac{R_{L} j ω C_{P} V_{o}}{1 + j ω R_{L} . (C_{P} + C_{L})}$

$At very high frequency$

$ω R_{L} (C_{P} + C_{L}) >> 1$

$∴ V_{L} = \frac{C_{P} V_{o}}{C_{P} + C_{L}}$

$B u t C_{P} V_{o} = Q (applied charge)$

$∴ V_{L} = \frac{Q}{C_{P} + C_{L}}$

$Given Q = 1nC,$

$∴ V_{L} = \frac{1 n C}{(1 + 0.5) n F} = 0.667 V$

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A piezoelectric transducer having capacitance of 1nF and leakage resistance of 1012 Ω is connected to a parallel load ZL. ZL comprises of 1 MΩ resistance (RL) and 0.5 nF capacitance (C1) in parallel. What will be the approximate value of load voltage VL at very high frequency for 1nC of charge?

The correct option is C 0.667 VThe load voltage is given as On Analysing we find, VL=RL jω CPVo1+jω RL.(CP+CL) At very high frequency ωRL(CP+CL)>>1 ∴VL=CPVoCP+CL But CPVo=Q(applied charge) ∴VL=QCP+CL Given Q = 1nC, ∴VL=1nC(1+0.5)nF=0.667 V