Question #61349

5) A galvanometer with coil resistance
12.0Ω
shows full scale deflection for a current of 2.5mA. How would you convert it into a voltmeter of range 0 to 10.0V?

a) 3988Ω
in series

b) 0.43Ω
in parallel

c) Ω
in parallel

d) 1.62Ω
in series

6) A series circuit consisting of an uncharged 42ÖF capacitor and
10MΩ
resistor is connected to 100V power source. What are the current in the circuit and the charge on the capacitor after one time constant?

a) 3.7μA
and
126μC

b) 4.6μA
and
221μC

c) 7.2μA
and
100μC

d) 1.3μA
and
52μC

Expert's answer

Answer on question #61349, Physics, Electric Circuits

5) A galvanometer with coil resistance 12.0Ω12.0\Omega shows full scale deflection for a current of 2.5mA2.5\mathrm{mA}. How would you convert it into a voltmeter of range 0 to 10.0V10.0\mathrm{V}?

a) 3988Ω3988\Omega in series

b) 0.43Ω0.43\Omega in parallel

c) Ω\Omega in parallel

d) 1.62Ω1.62\Omega in series

Solution:

According Ohms law


U=I(R+Rg)U = I (R + R _ {g})R+Rg=UIR + R _ {g} = \frac {U}{I}R=UI−RgR = \frac {U}{I} - R _ {g}R=10 V2.5⋅10−3 A−12.0 Ω=3988 ΩR = \frac {10\,V}{2.5 \cdot 10^{-3}\,A} - 12.0\,\Omega = 3988\,\Omega


**Answer:** a) 3988Ω3988\Omega in series

6) A series circuit consisting of an unchanged 42 nF capacitor and 10 MΩ resistor is connected to 100 V power source. What are the current in the circuit and the charge on the capacitor after one time constant?

a) 3.7μA3.7\mu\mathrm{A} and 126μC126\mu\mathrm{C}

b) 4.6μA4.6\mu\mathrm{A} and 221μC221\mu\mathrm{C}

c) 7.2μA7.2\mu\mathrm{A} and 100μC100\mu\mathrm{C}

d) 1.3μA1.3\mu\mathrm{A} and 52μC52\mu\mathrm{C}

Solution:

Current changes as


I=I0e−tτI = I _ {0} e ^ {- \frac {t}{\tau}}


Where t=τ=RCt = \tau = RC

The current is


I=I0e=VRe=1002.7⋅103=0.37⋅10−5 A=3.7 μAI = \frac {I _ {0}}{e} = \frac {V}{R e} = \frac {100}{2.7 \cdot 10^{3}} = 0.37 \cdot 10^{-5}\,A = 3.7\,\mu\mathrm{A}


The charge


q=q0e−tτ=CVe=100×42⋅10−122.7=1.56⋅10−9 C=156 μCq = q _ {0} e ^ {- \frac {t}{\tau}} = \frac {CV}{e} = \frac {100 \times 42 \cdot 10^{-12}}{2.7} = 1.56 \cdot 10^{-9}\,C = 156\,\mu\mathrm{C}


**Answer:** a) 3.7μA3.7\mu\mathrm{A} and 156μC156\mu\mathrm{C}

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