What is the maximum current that be drawn from a 1.50 - V battery with an intern

Question

What is the maximum current that be drawn from a 1.50 - V battery with an internal resistance of 0.30 ohm? 2.5A 4.5A 5.0 A 0.20 A 0.45 A For the circuit shown in the figure, all quantities are accurate to 3 significant figures. What is the power dissipated in the 2-ohm resistor? 3.56 W 5.33 W 2.67 W 6.67 W 8.0 W For the circuit shown in the figure, what is the current through resistor R_1? 0.13 A 0.071 A 0.029 A 0.016 A A 4.0-mF capacitor is discharged through a 4.0-k ohm resistor. How long will it take for the capacitor to lose half its initial charge? 2.7 s 5.5 s 9.2 s 1.1 s 1.6 s An electron, moving toward the west enters a uniform magnetic field. Because this field the electron curves upward. The direction of the magnetic field is towards the south. downward. downwards the north. upward. towards the west. An electron moving perpendicular to a uniform magnetic field of 3.2 times 10^-2 T moves in a circle of radius 0.40 cm. How fast is this electron moving? (m_r = 9.11 times 10^-33 kg.q_t = 1.60 times 10^-14 C) 1.9 times 10^-2 m/s 2.2 times 10^7 m/s 1.9 times 10^-30 m/s 8.0 times 10^6 m/s 3.0 times 10^6 m/s

Explanation / Answer

Q15 ans

Given that

voltage v=1.5 V

internal resistance R=0.3 ohm

now we find the current

current i=1.5/0.3=5 A

the correct option is C

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(Q16) ans

here we observe 2 ohm and 1 ohm are series connection

resistance R=2+1=3 ohm

resistance 5 and 1 ohms are series connnection

resistance R'=5+1=6 ohm

resistance R and R' are parallel connection

resistance R''=3*6/3+6=2 ohm

resistance R'' and 4 ohms are series connection

the equivalent resistance Req=2+4= 6 ohm

now we find the voltage on each resistance

voltage on resistance 4 ohm =>V2=4*12/4+2=8 v

voltage on resistance R'' ohm =>V1=2*12/2+4=4 v

voltage on 2 ohms resistance =>V1=4*2/2+1=2.67 v

now we find the power on 2 ohms

power P=2.67^2/2=3.56 W

correct option is A

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applying kirchhoff's first law

i1+i2=i3................................(1)

applying kirchhoff' second law

70 i1=7-2=5

current i1=0.071 A

the correct option B

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4 ans

Given that

capacitor C=4*10^-3 F

resistance R=4*10^3 ohm

now we find the time

Q/2=Q[e^-t/4*4]

log[0.5]=-t/16

time t=4.82 sec

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