Answer:
No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.
Explanation:
Answer:
Answered below.
Explanation:
A) Both spheroidite & tempered martensite possess sphere - like cementite particles within their microstructure known as a ferrite matrix. However, the difference is that these particles are much larger for spheroidite than tempered.
B) Tempered martensite is much harder and stronger than spheroidite primarily because there is much more ferrite - cementite phase boundary area for its sphere - like cementite particles.
This is because the greater the boundary area, the more the hardness.
Explanation:
A router is a switching device for networks, which is able to route network packets, based on their addresses, to other networks or devices. Among other things, they are used for Internet access, for coupling networks or for connecting branch offices to a central office via VPN (Virtual Private Network
Answer:
The operating principle of a ramp type digital voltmeter is to measure the time that a linear ramp voltage takes to change from level of input voltage to zero voltage (or vice versa).
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
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<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
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<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
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<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
