Latest ads

Baseus PI LITE INDOOR CAMERA
- ErMurazor
- Updated: Yesterday at 12:26 PM
Ad icon
Colombo Hard Disk For Sale
- mazza
- Updated: Saturday at 10:23 PM
land for sale
- justasking
- Updated: Saturday at 9:32 AM
AMD AM 5 Ryzen setup
- prabodhamadusanka
- Updated: Thursday at 12:56 PM
Gampaha ACER OKW223 Wired RGB Keyboard
- chint
- Updated: Thursday at 10:17 AM

Reply to thread

Message: <blockquote data-quote="AnuradhaRa" data-source="post: 31120930" data-attributes="member: 325076">Ah! Now I understand exactly what you’re asking — you are focusing on the cause-and-effect loop between current change and back EMF, independent of the applied voltage. Let’s clarify carefully.<hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="🔑" title="Key    :key:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/1f511.png" data-shortname=":key:" /> Key principle</h3>Back EMF (V_back) is caused by the rate of change of current:Vback=−LdIdtV_\text{back} = -L \frac{dI}{dt}Vback=−LdtdI<ul> <li data-xf-list-type="ul">If current is constant → no back EMF</li> <li data-xf-list-type="ul">If current changes → back EMF appears</li> <li data-xf-list-type="ul">Faster change → stronger back EMF</li> </ul>So you are right: the back EMF depends on dI/dt, not directly on voltage.<hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="🧩" title="Puzzle piece    :jigsaw:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/1f9e9.png" data-shortname=":jigsaw:" /> The “chicken-and-egg” question</h3>You’re asking:This is subtle but important.<hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="1️⃣" title="Keycap: 1    :one:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/0031-20e3.png" data-shortname=":one:" /> The initiating factor is applied voltage</h3>Even though back EMF depends on dI/dt, the rate of current change itself is forced by the applied voltage.From Kirchhoff’s law for an RL series circuit:Vapplied=IR+LdIdtV_\text{applied} = I R + L \frac{dI}{dt}Vapplied=IR+LdtdIRewriting:dIdt=Vapplied−IRL\frac{dI}{dt} = \frac{V_\text{applied} - I R}{L}dtdI=LVapplied−IR<ul> <li data-xf-list-type="ul">At t = 0, I = 0 → dIdt=VappliedL\frac{dI}{dt} = \frac{V_\text{applied}}{L}dtdI=LVapplied</li> <li data-xf-list-type="ul">So the applied voltage starts the current changing, and that change produces back EMF.</li> </ul>So:Applied voltage → causes initial dI/dt → produces back EMF → slows dI/dt<hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="2️⃣" title="Keycap: 2    :two:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/0032-20e3.png" data-shortname=":two:" /> The inductor resists change</h3> <ul> <li data-xf-list-type="ul">Once current starts rising, the inductor generates back EMF</li> <li data-xf-list-type="ul">This reduces the effective voltage pushing current: Veffective=Vapplied−VbackV_\text{effective} = V_\text{applied} - V_\text{back}Veffective=Vapplied−Vback</li> <li data-xf-list-type="ul">Therefore, dI/dt decreases over time</li> </ul>This is why the current rises gradually.<hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="⚡" title="High voltage    :zap:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/26a1.png" data-shortname=":zap:" /> Analogy</h3>Think of water in a tank:<ul> <li data-xf-list-type="ul">You open a faucet (applied voltage) → water starts flowing (current starts changing)</li> <li data-xf-list-type="ul">A valve resists flow proportionally to the flow rate (back EMF resists dI/dt)</li> <li data-xf-list-type="ul">Flow rises slowly until steady state</li> </ul><hr /><h3><img class="smilie smilie--emoji" loading="lazy" alt="✅" title="Check mark button    :white_check_mark:" src="https://cdn.jsdelivr.net/joypixels/assets/6.6/png/unicode/64/2705.png" data-shortname=":white_check_mark:" /> The subtle point</h3> <ul> <li data-xf-list-type="ul">Cause of back EMF: rate of current change (dI/dtdI/dtdI/dt)</li> <li data-xf-list-type="ul">Cause of current change in the first place: applied voltage</li> <li data-xf-list-type="ul">Effect: back EMF reduces rate of change</li> </ul>So yes, back EMF does not itself create current change; it resists it. The voltage source is what “forces” the initial change.<hr />If you want, I can draw a timeline diagram showing voltage, dI/dt, and back EMF together, which usually clears this confusion instantly.Do you want me to do that?</blockquote>

Verification: Awruddata maasa keeyada?

Top Bottom