question
a circuit contains an ammeter, a battery of 30 v and a resistance 40.8 ohm all connected in series. if the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:



 option 1)
 0.25 a
 
 option 2)
 2 a
 
 option 3)
 1 a
 
 option 4)
 0.5 a
472
(a) material q with resistivity 2.63×10-8 ohm-m can be used for making electric wires because it has very low resistivity.(b) material r with resistivity 1.0 ×1015 ohm-m can be used for making handle of soldering iron because it has very high resistivity.(c) material p with resistivity 2.3 × 103 ohm-m can be used for making solar cell because it is a semiconductor. from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] (i) here current (i) = 1 a, resistance (r) = 300 ohm, time(t) = 1hwe know thatp = i2r = 12 x 300 = 300 we = p x t = 300 x 1 = 300 wh(ii) here current (i) = 2 a resistance(r) = 100 ohm, time (t) = 1hwe know thatp = i2r = 22 x 100 = 400 we = p x t = 400 x 1 = 400 whhence, in case (ii), the electrical energy consumed per hour is more. ![How do you test a coil?](https://static.3dstockphoto.com/thumb/00/00/61/90/uwx619072.jpg)  
Photo Credit: Courtesy of 3DStockPhoto ([Do You image](https://3dstockphoto.com/Do-You-uwx619072))The only safe way to test for spark is to use a spark plug tester tool. If a coil problem is suspected, measure the coil's primary and secondary resistance with an ohmmeter. If either is out of specifications, the coil needs to be replaced. A coil can be easily bench tested with a digital 10 megaohm impedance ohmmeter. if you right click on the battery\_simple\_dynamics subsystem, you can go to mask -> look under mask. here, you will see that the state of charge is being computed in this subsystem. the ah*0.2*3600 constant block after changing to ah*0.1*3600 i then see an initial charge state of 90%. let a and b be the two given vectorsɵ = 150° is the angle between a and bif c is the resultant vector, then: ...1also, it is given that the resultant is perpendicular to one of the given vectors. let that given vector be anow, …2putting 2 in 1, we get:now, putting c = 10, ɵ = 150° ()so, hence, smaller vector is a = 20 units All that is needed is some sort of wave-shape that sweeps the oscilloscope beam to and fro. Since the relationship between vertical displacement and horizontal displacement on the oscilloscope’s trace is purely a function of the test component’s characteristic curve, we really don’t care what form of [AC](https://www.allaboutcircuits.com/textbook/alternating-current/chpt-1/what-is-alternating-current-ac/) voltage excites it. we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> as the length of the wire increases so does the electrical resistance as electrons experience a smaller pd gradient as wire gets longer. the size of voltage per metre value decreases causing a lower drift velocity and more ions for electrons to collide with